Termination w.r.t. Q proof of AProVE_09_Inductive

Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS

↳1 Overlay + Local Confluence (⇔, 0 ms)

↳2 QTRS

↳3 DependencyPairsProof (⇔, 0 ms)

↳4 QDP

↳5 DependencyGraphProof (⇔, 0 ms)

↳6 AND

    ↳7 QDP

        ↳8 UsableRulesProof (⇔, 0 ms)

        ↳9 QDP

        ↳10 QReductionProof (⇔, 0 ms)

        ↳11 QDP

        ↳12 QDPSizeChangeProof (⇔, 0 ms)

        ↳13 YES

    ↳14 QDP

        ↳15 UsableRulesProof (⇔, 0 ms)

        ↳16 QDP

        ↳17 QReductionProof (⇔, 0 ms)

        ↳18 QDP

        ↳19 QDPSizeChangeProof (⇔, 0 ms)

        ↳20 YES

    ↳21 QDP

        ↳22 UsableRulesProof (⇔, 0 ms)

        ↳23 QDP

        ↳24 QReductionProof (⇔, 0 ms)

        ↳25 QDP

        ↳26 QDPSizeChangeProof (⇔, 0 ms)

        ↳27 YES

    ↳28 QDP

        ↳29 UsableRulesProof (⇔, 0 ms)

        ↳30 QDP

        ↳31 QReductionProof (⇔, 0 ms)

        ↳32 QDP

        ↳33 QDPOrderProof (⇔, 0 ms)

        ↳34 QDP

        ↳35 PisEmptyProof (⇔, 0 ms)

        ↳36 YES

    ↳37 QDP

        ↳38 UsableRulesProof (⇔, 0 ms)

        ↳39 QDP

        ↳40 QReductionProof (⇔, 0 ms)

        ↳41 QDP

        ↳42 TransformationProof (⇔, 2 ms)

        ↳43 QDP

        ↳44 DependencyGraphProof (⇔, 0 ms)

        ↳45 QDP

        ↳46 UsableRulesProof (⇔, 0 ms)

        ↳47 QDP

        ↳48 QReductionProof (⇔, 0 ms)

        ↳49 QDP

        ↳50 TransformationProof (⇔, 0 ms)

        ↳51 QDP

        ↳52 Induction-Processor (⇒, 45 ms)

        ↳53 AND

            ↳54 QDP

                ↳55 DependencyGraphProof (⇔, 0 ms)

                ↳56 TRUE

            ↳57 QTRS

                ↳58 Overlay + Local Confluence (⇔, 0 ms)

                ↳59 QTRS

                ↳60 DependencyPairsProof (⇔, 0 ms)

                ↳61 QDP

                ↳62 DependencyGraphProof (⇔, 0 ms)

                ↳63 AND

                    ↳64 QDP

                        ↳65 UsableRulesProof (⇔, 0 ms)

                        ↳66 QDP

                        ↳67 QReductionProof (⇔, 0 ms)

                        ↳68 QDP

                        ↳69 QDPSizeChangeProof (⇔, 0 ms)

                        ↳70 YES

                    ↳71 QDP

                        ↳72 UsableRulesProof (⇔, 0 ms)

                        ↳73 QDP

                        ↳74 QReductionProof (⇔, 0 ms)

                        ↳75 QDP

                        ↳76 QDPSizeChangeProof (⇔, 0 ms)

                        ↳77 YES

                    ↳78 QDP

                        ↳79 UsableRulesProof (⇔, 0 ms)

                        ↳80 QDP

                        ↳81 QReductionProof (⇔, 0 ms)

                        ↳82 QDP

                        ↳83 QDPSizeChangeProof (⇔, 0 ms)

                        ↳84 YES

                    ↳85 QDP

                        ↳86 UsableRulesProof (⇔, 0 ms)

                        ↳87 QDP

                        ↳88 QReductionProof (⇔, 0 ms)

                        ↳89 QDP

                        ↳90 QDPSizeChangeProof (⇔, 0 ms)

                        ↳91 YES

                    ↳92 QDP

                        ↳93 UsableRulesProof (⇔, 0 ms)

                        ↳94 QDP

                        ↳95 QReductionProof (⇔, 0 ms)

                        ↳96 QDP

                        ↳97 QDPSizeChangeProof (⇔, 0 ms)

                        ↳98 YES

                    ↳99 QDP

                        ↳100 UsableRulesProof (⇔, 0 ms)

                        ↳101 QDP

                        ↳102 QReductionProof (⇔, 0 ms)

                        ↳103 QDP

                        ↳104 QDPOrderProof (⇔, 0 ms)

                        ↳105 QDP

                        ↳106 PisEmptyProof (⇔, 0 ms)

                        ↳107 YES

                    ↳108 QDP

                        ↳109 UsableRulesProof (⇔, 0 ms)

                        ↳110 QDP

                        ↳111 QReductionProof (⇔, 0 ms)

                        ↳112 QDP

                        ↳113 QDPSizeChangeProof (⇔, 0 ms)

                        ↳114 YES

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
sort(xs) → if3(empty(xs), xs)
if3(true, xs) → nil
if3(false, xs) → sort(del(max(xs), xs))
empty(nil) → true
empty(cons(x, xs)) → false
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
sort(xs) → if3(empty(xs), xs)
if3(true, xs) → nil
if3(false, xs) → sort(del(max(xs), xs))
empty(nil) → true
empty(cons(x, xs)) → false
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MAX(cons(x, cons(y, xs))) → IF1(ge(x, y), x, y, xs)
MAX(cons(x, cons(y, xs))) → GE(x, y)
IF1(true, x, y, xs) → MAX(cons(x, xs))
IF1(false, x, y, xs) → MAX(cons(y, xs))
DEL(x, cons(y, xs)) → IF2(eq(x, y), x, y, xs)
DEL(x, cons(y, xs)) → EQ(x, y)
IF2(false, x, y, xs) → DEL(x, xs)
EQ(s(x), s(y)) → EQ(x, y)
SORT(xs) → IF3(empty(xs), xs)
SORT(xs) → EMPTY(xs)
IF3(false, xs) → SORT(del(max(xs), xs))
IF3(false, xs) → DEL(max(xs), xs)
IF3(false, xs) → MAX(xs)
GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
sort(xs) → if3(empty(xs), xs)
if3(true, xs) → nil
if3(false, xs) → sort(del(max(xs), xs))
empty(nil) → true
empty(cons(x, xs)) → false
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 5 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
sort(xs) → if3(empty(xs), xs)
if3(true, xs) → nil
if3(false, xs) → sort(del(max(xs), xs))
empty(nil) → true
empty(cons(x, xs)) → false
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(8) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

Q DP problem:
The TRS P consists of the following rules:

GE(s(x), s(y)) → GE(x, y)

R is empty.
The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(10) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

(11) Obligation:

Q DP problem:
The TRS P consists of the following rules:

GE(s(x), s(y)) → GE(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

GE(s(x), s(y)) → GE(x, y)
The graph contains the following edges 1 > 1, 2 > 2

(13) YES

(14) Obligation:

Q DP problem:
The TRS P consists of the following rules:

EQ(s(x), s(y)) → EQ(x, y)

The TRS R consists of the following rules:

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
sort(xs) → if3(empty(xs), xs)
if3(true, xs) → nil
if3(false, xs) → sort(del(max(xs), xs))
empty(nil) → true
empty(cons(x, xs)) → false
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(15) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

Q DP problem:
The TRS P consists of the following rules:

EQ(s(x), s(y)) → EQ(x, y)

R is empty.
The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

(18) Obligation:

Q DP problem:
The TRS P consists of the following rules:

EQ(s(x), s(y)) → EQ(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(19) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

EQ(s(x), s(y)) → EQ(x, y)
The graph contains the following edges 1 > 1, 2 > 2

(20) YES

(21) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF2(false, x, y, xs) → DEL(x, xs)
DEL(x, cons(y, xs)) → IF2(eq(x, y), x, y, xs)

The TRS R consists of the following rules:

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
sort(xs) → if3(empty(xs), xs)
if3(true, xs) → nil
if3(false, xs) → sort(del(max(xs), xs))
empty(nil) → true
empty(cons(x, xs)) → false
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(22) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(23) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF2(false, x, y, xs) → DEL(x, xs)
DEL(x, cons(y, xs)) → IF2(eq(x, y), x, y, xs)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(24) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
sort(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

(25) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF2(false, x, y, xs) → DEL(x, xs)
DEL(x, cons(y, xs)) → IF2(eq(x, y), x, y, xs)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(26) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

DEL(x, cons(y, xs)) → IF2(eq(x, y), x, y, xs)
The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4
IF2(false, x, y, xs) → DEL(x, xs)
The graph contains the following edges 2 >= 1, 4 >= 2

(27) YES

(28) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF1(true, x, y, xs) → MAX(cons(x, xs))
MAX(cons(x, cons(y, xs))) → IF1(ge(x, y), x, y, xs)
IF1(false, x, y, xs) → MAX(cons(y, xs))

The TRS R consists of the following rules:

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
sort(xs) → if3(empty(xs), xs)
if3(true, xs) → nil
if3(false, xs) → sort(del(max(xs), xs))
empty(nil) → true
empty(cons(x, xs)) → false
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(29) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(30) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF1(true, x, y, xs) → MAX(cons(x, xs))
MAX(cons(x, cons(y, xs))) → IF1(ge(x, y), x, y, xs)
IF1(false, x, y, xs) → MAX(cons(y, xs))

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(31) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))

(32) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF1(true, x, y, xs) → MAX(cons(x, xs))
MAX(cons(x, cons(y, xs))) → IF1(ge(x, y), x, y, xs)
IF1(false, x, y, xs) → MAX(cons(y, xs))

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(33) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

IF1(true, x, y, xs) → MAX(cons(x, xs))
MAX(cons(x, cons(y, xs))) → IF1(ge(x, y), x, y, xs)
IF1(false, x, y, xs) → MAX(cons(y, xs))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
IF1(x1, x2, x3, x4) = IF1(x4)
MAX(x1) = x1
cons(x1, x2) = cons(x2)

Knuth-Bendix order [KBO] with precedence:

trivial

and weight map:

dummyConstant=1
IF1_1=3
cons_1=2

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none

(34) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(35) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(36) YES

(37) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF3(false, xs) → SORT(del(max(xs), xs))
SORT(xs) → IF3(empty(xs), xs)

The TRS R consists of the following rules:

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
sort(xs) → if3(empty(xs), xs)
if3(true, xs) → nil
if3(false, xs) → sort(del(max(xs), xs))
empty(nil) → true
empty(cons(x, xs)) → false
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(38) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(39) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF3(false, xs) → SORT(del(max(xs), xs))
SORT(xs) → IF3(empty(xs), xs)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, xs)) → false
max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(40) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

sort(x0)
if3(true, x0)
if3(false, x0)

(41) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF3(false, xs) → SORT(del(max(xs), xs))
SORT(xs) → IF3(empty(xs), xs)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, xs)) → false
max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(42) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule SORT(xs) → IF3(empty(xs), xs) at position [0] we obtained the following new rules [LPAR04]:

SORT(nil) → IF3(true, nil) → SORT(nil) → IF3(true, nil)
SORT(cons(x0, x1)) → IF3(false, cons(x0, x1)) → SORT(cons(x0, x1)) → IF3(false, cons(x0, x1))

(43) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF3(false, xs) → SORT(del(max(xs), xs))
SORT(nil) → IF3(true, nil)
SORT(cons(x0, x1)) → IF3(false, cons(x0, x1))

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, xs)) → false
max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(44) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(45) Obligation:

Q DP problem:
The TRS P consists of the following rules:

SORT(cons(x0, x1)) → IF3(false, cons(x0, x1))
IF3(false, xs) → SORT(del(max(xs), xs))

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, xs)) → false
max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(46) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(47) Obligation:

Q DP problem:
The TRS P consists of the following rules:

SORT(cons(x0, x1)) → IF3(false, cons(x0, x1))
IF3(false, xs) → SORT(del(max(xs), xs))

The TRS R consists of the following rules:

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(48) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

empty(nil)
empty(cons(x0, x1))

(49) Obligation:

Q DP problem:
The TRS P consists of the following rules:

SORT(cons(x0, x1)) → IF3(false, cons(x0, x1))
IF3(false, xs) → SORT(del(max(xs), xs))

The TRS R consists of the following rules:

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(50) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF3(false, xs) → SORT(del(max(xs), xs)) we obtained the following new rules [LPAR04]:

IF3(false, cons(z0, z1)) → SORT(del(max(cons(z0, z1)), cons(z0, z1))) → IF3(false, cons(z0, z1)) → SORT(del(max(cons(z0, z1)), cons(z0, z1)))

(51) Obligation:

Q DP problem:
The TRS P consists of the following rules:

SORT(cons(x0, x1)) → IF3(false, cons(x0, x1))
IF3(false, cons(z0, z1)) → SORT(del(max(cons(z0, z1)), cons(z0, z1)))

The TRS R consists of the following rules:

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(52) Induction-Processor (SOUND transformation)

This DP could be deleted by the Induction-Processor:
IF3(false_renamed, cons(z0, z1)) → SORT(del(max(cons(z0, z1)), cons(z0, z1)))

This order was computed:
Polynomial interpretation [POLO]:

POL(0) = 1
POL(IF3(x₁, x₂)) = x₂
POL(SORT(x₁)) = x₁
POL(cons(x₁, x₂)) = 1 + x₁ + x₂
POL(del(x₁, x₂)) = x₂
POL(eq(x₁, x₂)) = 1 + x₁ + x₂
POL(false_renamed) = 0
POL(ge(x₁, x₂)) = 1
POL(if1(x₁, x₂, x₃, x₄)) = 1 + x₂ + x₃ + x₄
POL(if2(x₁, x₂, x₃, x₄)) = 1 + x₃ + x₄
POL(max(x₁)) = x₁
POL(nil) = 1
POL(s(x₁)) = x₁
POL(true_renamed) = 1

At least one of these decreasing rules is always used after the deleted DP:
if2(true_renamed, x7, y4, xs2) → xs2

The following formula is valid:
z2:sort[a34].(¬(z2 =nil)→del'(max(z2 ), z2 )=true)

The transformed set:
del'(x3, nil) → false
del'(x4, cons(y1, xs1)) → if2'(eq(x4, y1), x4, y1, xs1)
if2'(true_renamed, x7, y4, xs2) → true
if2'(false_renamed, x8, y5, xs3) → del'(x8, xs3)
max(nil) → 0
max(cons(x, nil)) → x
max(cons(x', cons(y, xs))) → if1(ge(x', y), x', y, xs)
if1(true_renamed, x'', y', xs') → max(cons(x'', xs'))
if1(false_renamed, x2, y'', xs'') → max(cons(y'', xs''))
del(x3, nil) → nil
del(x4, cons(y1, xs1)) → if2(eq(x4, y1), x4, y1, xs1)
eq(0, 0) → true_renamed
eq(0, s(y2)) → false_renamed
eq(s(x5), 0) → false_renamed
eq(s(x6), s(y3)) → eq(x6, y3)
if2(true_renamed, x7, y4, xs2) → xs2
if2(false_renamed, x8, y5, xs3) → cons(y5, del(x8, xs3))
ge(x9, 0) → true_renamed
ge(0, s(x10)) → false_renamed
ge(s(x11), s(y6)) → ge(x11, y6)
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a33](0, 0) → true
equal_sort[a33](0, s(v54)) → false
equal_sort[a33](s(v55), 0) → false
equal_sort[a33](s(v55), s(v56)) → equal_sort[a33](v55, v56)
equal_sort[a34](nil, nil) → true
equal_sort[a34](nil, cons(v57, v58)) → false
equal_sort[a34](cons(v59, v60), nil) → false
equal_sort[a34](cons(v59, v60), cons(v61, v62)) → and(equal_sort[a33](v59, v61), equal_sort[a34](v60, v62))
equal_sort[a46](true_renamed, true_renamed) → true
equal_sort[a46](true_renamed, false_renamed) → false
equal_sort[a46](false_renamed, true_renamed) → false
equal_sort[a46](false_renamed, false_renamed) → true
equal_sort[a63](witness_sort[a63], witness_sort[a63]) → true

The proof given by the theorem prover:
The following output was given by the internal theorem prover:

proof of internal
# AProVE Commit ID: 3a20a6ef7432c3f292db1a8838479c42bf5e3b22 root 20240618 unpublished


Partial correctness of the following Program

   [x, v54, v55, v56, v57, v58, v59, v60, v61, v62, x3, x4, y1, xs1, x7, y4, xs2, x8, y5, x', y, xs, y2, x5, x6, y3, x9, x10, x11, y6, x'', y', x2, y'', xs', xs'']
   equal_bool(true, false) -> false
   equal_bool(false, true) -> false
   equal_bool(true, true) -> true
   equal_bool(false, false) -> true
   true and x -> x
   false and x -> false
   true or x -> true
   false or x -> x
   not(false) -> true
   not(true) -> false
   isa_true(true) -> true
   isa_true(false) -> false
   isa_false(true) -> false
   isa_false(false) -> true
   equal_sort[a33](0, 0) -> true
   equal_sort[a33](0, s(v54)) -> false
   equal_sort[a33](s(v55), 0) -> false
   equal_sort[a33](s(v55), s(v56)) -> equal_sort[a33](v55, v56)
   equal_sort[a34](nil, nil) -> true
   equal_sort[a34](nil, cons(v57, v58)) -> false
   equal_sort[a34](cons(v59, v60), nil) -> false
   equal_sort[a34](cons(v59, v60), cons(v61, v62)) -> equal_sort[a33](v59, v61) and equal_sort[a34](v60, v62)
   equal_sort[a46](true_renamed, true_renamed) -> true
   equal_sort[a46](true_renamed, false_renamed) -> false
   equal_sort[a46](false_renamed, true_renamed) -> false
   equal_sort[a46](false_renamed, false_renamed) -> true
   equal_sort[a63](witness_sort[a63], witness_sort[a63]) -> true
   del'(x3, nil) -> false
   equal_sort[a46](eq(x4, y1), true_renamed) -> true | del'(x4, cons(y1, xs1)) -> true
   equal_sort[a46](eq(x4, y1), true_renamed) -> false | del'(x4, cons(y1, xs1)) -> del'(x4, xs1)
   if2'(true_renamed, x7, y4, xs2) -> true
   if2'(false_renamed, x8, y5, nil) -> false
   if2'(false_renamed, x8, y5, cons(y1, xs1)) -> if2'(eq(x8, y1), x8, y1, xs1)
   max(nil) -> 0
   max(cons(x, nil)) -> x
   equal_sort[a46](ge(x', y), true_renamed) -> true | max(cons(x', cons(y, xs))) -> max(cons(x', xs))
   equal_sort[a46](ge(x', y), true_renamed) -> false | max(cons(x', cons(y, xs))) -> max(cons(y, xs))
   del(x3, nil) -> nil
   equal_sort[a46](eq(x4, y1), true_renamed) -> true | del(x4, cons(y1, xs1)) -> xs1
   equal_sort[a46](eq(x4, y1), true_renamed) -> false | del(x4, cons(y1, xs1)) -> cons(y1, del(x4, xs1))
   eq(0, 0) -> true_renamed
   eq(0, s(y2)) -> false_renamed
   eq(s(x5), 0) -> false_renamed
   eq(s(x6), s(y3)) -> eq(x6, y3)
   if2(true_renamed, x7, y4, xs2) -> xs2
   if2(false_renamed, x8, y5, nil) -> cons(y5, nil)
   if2(false_renamed, x8, y5, cons(y1, xs1)) -> cons(y5, if2(eq(x8, y1), x8, y1, xs1))
   ge(x9, 0) -> true_renamed
   ge(0, s(x10)) -> false_renamed
   ge(s(x11), s(y6)) -> ge(x11, y6)
   if1(true_renamed, x'', y', nil) -> x''
   if1(true_renamed, x'', y', cons(y, xs)) -> if1(ge(x'', y), x'', y, xs)
   if1(false_renamed, x2, y'', nil) -> y''
   if1(false_renamed, x2, y'', cons(y, xs)) -> if1(ge(y'', y), y'', y, xs)
   if1(true_renamed, x'', y', xs') -> 0
   if1(false_renamed, x2, y'', xs'') -> 0

using the following formula:
z2:sort[a34].(~(z2=nil)->del'(max(z2), z2)=true)

could be successfully shown:
(0) Formula
(1) Induction by algorithm [EQUIVALENT, 0 ms]
(2) AND
    (3) Formula
        (4) Symbolic evaluation [EQUIVALENT, 0 ms]
        (5) YES
    (6) Formula
        (7) Symbolic evaluation [EQUIVALENT, 0 ms]
        (8) Formula
        (9) Induction by data structure [EQUIVALENT, 0 ms]
        (10) AND
            (11) Formula
                (12) Symbolic evaluation [EQUIVALENT, 0 ms]
                (13) YES
            (14) Formula
                (15) Conditional Evaluation [EQUIVALENT, 0 ms]
                (16) AND
                    (17) Formula
                        (18) Symbolic evaluation [EQUIVALENT, 0 ms]
                        (19) YES
                    (20) Formula
                        (21) Symbolic evaluation [EQUIVALENT, 0 ms]
                        (22) Formula
                        (23) Hypothesis Lifting [EQUIVALENT, 0 ms]
                        (24) Formula
                        (25) Symbolic evaluation under hypothesis [SOUND, 0 ms]
                        (26) Formula
                        (27) Hypothesis Lifting [EQUIVALENT, 0 ms]
                        (28) Formula
                        (29) Hypothesis Lifting [EQUIVALENT, 0 ms]
                        (30) Formula
                        (31) Conditional Evaluation [EQUIVALENT, 0 ms]
                        (32) AND
                            (33) Formula
                                (34) Symbolic evaluation under hypothesis [EQUIVALENT, 0 ms]
                                (35) YES
                            (36) Formula
                                (37) Symbolic evaluation [EQUIVALENT, 0 ms]
                                (38) YES
    (39) Formula
        (40) Symbolic evaluation [EQUIVALENT, 0 ms]
        (41) Formula
        (42) Conditional Evaluation [EQUIVALENT, 0 ms]
        (43) Formula
        (44) Conditional Evaluation [EQUIVALENT, 0 ms]
        (45) AND
            (46) Formula
                (47) Symbolic evaluation [EQUIVALENT, 0 ms]
                (48) YES
            (49) Formula
                (50) Conditional Evaluation [EQUIVALENT, 0 ms]
                (51) AND
                    (52) Formula
                        (53) Symbolic evaluation [EQUIVALENT, 0 ms]
                        (54) YES
                    (55) Formula
                        (56) Hypothesis Lifting [EQUIVALENT, 0 ms]
                        (57) Formula
                        (58) Conditional Evaluation [EQUIVALENT, 0 ms]
                        (59) Formula
                        (60) Symbolic evaluation [EQUIVALENT, 0 ms]
                        (61) YES
    (62) Formula
        (63) Symbolic evaluation [EQUIVALENT, 0 ms]
        (64) Formula
        (65) Conditional Evaluation [EQUIVALENT, 0 ms]
        (66) Formula
        (67) Conditional Evaluation [EQUIVALENT, 0 ms]
        (68) AND
            (69) Formula
                (70) Symbolic evaluation [EQUIVALENT, 0 ms]
                (71) YES
            (72) Formula
                (73) Symbolic evaluation under hypothesis [EQUIVALENT, 0 ms]
                (74) YES


----------------------------------------

(0)
Obligation:
Formula:
z2:sort[a34].(~(z2=nil)->del'(max(z2), z2)=true)

There are no hypotheses.




----------------------------------------

(1) Induction by algorithm (EQUIVALENT)
Induction by algorithm max(z2) generates the following cases:

1. Base Case:
Formula:
(~(nil=nil)->del'(max(nil), nil)=true)

There are no hypotheses.





2. Base Case:
Formula:
x:sort[a33].(~(cons(x, nil)=nil)->del'(max(cons(x, nil)), cons(x, nil))=true)

There are no hypotheses.





1. Step Case:
Formula:
x':sort[a33],y:sort[a33],xs:sort[a34].(~(cons(x', cons(y, xs))=nil)->del'(max(cons(x', cons(y, xs))), cons(x', cons(y, xs)))=true)

Hypotheses:
x':sort[a33],xs:sort[a34].del'(max(cons(x', xs)), cons(x', xs))=true
x':sort[a33],y:sort[a33].equal_sort[a46](ge(x', y), true_renamed)=true





2. Step Case:
Formula:
x':sort[a33],y:sort[a33],xs:sort[a34].(~(cons(x', cons(y, xs))=nil)->del'(max(cons(x', cons(y, xs))), cons(x', cons(y, xs)))=true)

Hypotheses:
y:sort[a33],xs:sort[a34].del'(max(cons(y, xs)), cons(y, xs))=true
x':sort[a33],y:sort[a33].equal_sort[a46](ge(x', y), true_renamed)=false






----------------------------------------

(2)
Complex Obligation (AND)

----------------------------------------

(3)
Obligation:
Formula:
(~(nil=nil)->del'(max(nil), nil)=true)

There are no hypotheses.




----------------------------------------

(4) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(5)
YES

----------------------------------------

(6)
Obligation:
Formula:
x:sort[a33].(~(cons(x, nil)=nil)->del'(max(cons(x, nil)), cons(x, nil))=true)

There are no hypotheses.




----------------------------------------

(7) Symbolic evaluation (EQUIVALENT)
Could be shown by simple symbolic evaluation.
----------------------------------------

(8)
Obligation:
Formula:
x:sort[a33].del'(x, cons(x, nil))=true

There are no hypotheses.




----------------------------------------

(9) Induction by data structure (EQUIVALENT)
Induction by data structure sort[a33] generates the following cases:



1. Base Case:
Formula:
del'(0, cons(0, nil))=true

There are no hypotheses.





1. Step Case:
Formula:
n:sort[a33].del'(s(n), cons(s(n), nil))=true

Hypotheses:
n:sort[a33].del'(n, cons(n, nil))=true






----------------------------------------

(10)
Complex Obligation (AND)

----------------------------------------

(11)
Obligation:
Formula:
del'(0, cons(0, nil))=true

There are no hypotheses.




----------------------------------------

(12) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(13)
YES

----------------------------------------

(14)
Obligation:
Formula:
n:sort[a33].del'(s(n), cons(s(n), nil))=true

Hypotheses:
n:sort[a33].del'(n, cons(n, nil))=true




----------------------------------------

(15) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
true=true

Hypotheses:
n:sort[a33].del'(n, cons(n, nil))=true
n:sort[a33].equal_sort[a46](eq(s(n), s(n)), true_renamed)=true





Formula:
n:sort[a33].del'(s(n), nil)=true

Hypotheses:
n:sort[a33].del'(n, cons(n, nil))=true
n:sort[a33].equal_sort[a46](eq(s(n), s(n)), true_renamed)=false






----------------------------------------

(16)
Complex Obligation (AND)

----------------------------------------

(17)
Obligation:
Formula:
true=true

Hypotheses:
n:sort[a33].del'(n, cons(n, nil))=true
n:sort[a33].equal_sort[a46](eq(s(n), s(n)), true_renamed)=true




----------------------------------------

(18) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(19)
YES

----------------------------------------

(20)
Obligation:
Formula:
n:sort[a33].del'(s(n), nil)=true

Hypotheses:
n:sort[a33].del'(n, cons(n, nil))=true
n:sort[a33].equal_sort[a46](eq(s(n), s(n)), true_renamed)=false




----------------------------------------

(21) Symbolic evaluation (EQUIVALENT)
Could be shown by simple symbolic evaluation.
----------------------------------------

(22)
Obligation:
Formula:
False

Hypotheses:
n:sort[a33].del'(n, cons(n, nil))=true
n:sort[a33].equal_sort[a46](eq(s(n), s(n)), true_renamed)=false




----------------------------------------

(23) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
n:sort[a33].((del'(n, cons(n, nil))=true/\equal_sort[a46](eq(s(n), s(n)), true_renamed)=false)->False)

Hypotheses:
n:sort[a33].del'(n, cons(n, nil))=true
n:sort[a33].equal_sort[a46](eq(s(n), s(n)), true_renamed)=false




----------------------------------------

(24)
Obligation:
Formula:
n:sort[a33].((del'(n, cons(n, nil))=true/\equal_sort[a46](eq(s(n), s(n)), true_renamed)=false)->False)

Hypotheses:
n:sort[a33].del'(n, cons(n, nil))=true
n:sort[a33].equal_sort[a46](eq(s(n), s(n)), true_renamed)=false




----------------------------------------

(25) Symbolic evaluation under hypothesis (SOUND)
Could be reduced by symbolic evaluation under hypothesis to:
n:sort[a33].~(equal_sort[a46](eq(n, n), true_renamed)=false)

By using the following hypotheses:
n:sort[a33].del'(n, cons(n, nil))=true

----------------------------------------

(26)
Obligation:
Formula:
n:sort[a33].~(equal_sort[a46](eq(n, n), true_renamed)=false)

Hypotheses:
n:sort[a33].del'(n, cons(n, nil))=true
n:sort[a33].equal_sort[a46](eq(s(n), s(n)), true_renamed)=false




----------------------------------------

(27) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
n:sort[a33].(equal_sort[a46](eq(n, n), true_renamed)=false->~(equal_sort[a46](eq(n, n), true_renamed)=false))

Hypotheses:
n:sort[a33].del'(n, cons(n, nil))=true




----------------------------------------

(28)
Obligation:
Formula:
n:sort[a33].(equal_sort[a46](eq(n, n), true_renamed)=false->~(equal_sort[a46](eq(n, n), true_renamed)=false))

Hypotheses:
n:sort[a33].del'(n, cons(n, nil))=true




----------------------------------------

(29) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
n:sort[a33].(del'(n, cons(n, nil))=true->(equal_sort[a46](eq(n, n), true_renamed)=false->~(equal_sort[a46](eq(n, n), true_renamed)=false)))

There are no hypotheses.




----------------------------------------

(30)
Obligation:
Formula:
n:sort[a33].(del'(n, cons(n, nil))=true->(equal_sort[a46](eq(n, n), true_renamed)=false->~(equal_sort[a46](eq(n, n), true_renamed)=false)))

There are no hypotheses.




----------------------------------------

(31) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
n:sort[a33].(true=true->(equal_sort[a46](eq(n, n), true_renamed)=false->~(equal_sort[a46](eq(n, n), true_renamed)=false)))

Hypotheses:
n:sort[a33].equal_sort[a46](eq(n, n), true_renamed)=true





Formula:
n:sort[a33].(del'(n, nil)=true->(equal_sort[a46](eq(n, n), true_renamed)=false->~(equal_sort[a46](eq(n, n), true_renamed)=false)))

Hypotheses:
n:sort[a33].equal_sort[a46](eq(n, n), true_renamed)=false






----------------------------------------

(32)
Complex Obligation (AND)

----------------------------------------

(33)
Obligation:
Formula:
n:sort[a33].(true=true->(equal_sort[a46](eq(n, n), true_renamed)=false->~(equal_sort[a46](eq(n, n), true_renamed)=false)))

Hypotheses:
n:sort[a33].equal_sort[a46](eq(n, n), true_renamed)=true




----------------------------------------

(34) Symbolic evaluation under hypothesis (EQUIVALENT)
Could be shown using symbolic evaluation under hypothesis, by using the following hypotheses:

n:sort[a33].equal_sort[a46](eq(n, n), true_renamed)=true

----------------------------------------

(35)
YES

----------------------------------------

(36)
Obligation:
Formula:
n:sort[a33].(del'(n, nil)=true->(equal_sort[a46](eq(n, n), true_renamed)=false->~(equal_sort[a46](eq(n, n), true_renamed)=false)))

Hypotheses:
n:sort[a33].equal_sort[a46](eq(n, n), true_renamed)=false




----------------------------------------

(37) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(38)
YES

----------------------------------------

(39)
Obligation:
Formula:
x':sort[a33],y:sort[a33],xs:sort[a34].(~(cons(x', cons(y, xs))=nil)->del'(max(cons(x', cons(y, xs))), cons(x', cons(y, xs)))=true)

Hypotheses:
x':sort[a33],xs:sort[a34].del'(max(cons(x', xs)), cons(x', xs))=true
x':sort[a33],y:sort[a33].equal_sort[a46](ge(x', y), true_renamed)=true




----------------------------------------

(40) Symbolic evaluation (EQUIVALENT)
Could be shown by simple symbolic evaluation.
----------------------------------------

(41)
Obligation:
Formula:
x':sort[a33],y:sort[a33],xs:sort[a34].del'(max(cons(x', cons(y, xs))), cons(x', cons(y, xs)))=true

Hypotheses:
x':sort[a33],xs:sort[a34].del'(max(cons(x', xs)), cons(x', xs))=true
x':sort[a33],y:sort[a33].equal_sort[a46](ge(x', y), true_renamed)=true




----------------------------------------

(42) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
x':sort[a33],xs:sort[a34],y:sort[a33].del'(max(cons(x', xs)), cons(x', cons(y, xs)))=true

Hypotheses:
x':sort[a33],xs:sort[a34].del'(max(cons(x', xs)), cons(x', xs))=true
x':sort[a33],y:sort[a33].equal_sort[a46](ge(x', y), true_renamed)=true






----------------------------------------

(43)
Obligation:
Formula:
x':sort[a33],xs:sort[a34],y:sort[a33].del'(max(cons(x', xs)), cons(x', cons(y, xs)))=true

Hypotheses:
x':sort[a33],xs:sort[a34].del'(max(cons(x', xs)), cons(x', xs))=true
x':sort[a33],y:sort[a33].equal_sort[a46](ge(x', y), true_renamed)=true




----------------------------------------

(44) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
true=true

Hypotheses:
x':sort[a33],xs:sort[a34].del'(max(cons(x', xs)), cons(x', xs))=true
x':sort[a33],y:sort[a33].equal_sort[a46](ge(x', y), true_renamed)=true
x':sort[a33],xs:sort[a34].equal_sort[a46](eq(max(cons(x', xs)), x'), true_renamed)=true





Formula:
x':sort[a33],xs:sort[a34],y:sort[a33].del'(max(cons(x', xs)), cons(y, xs))=true

Hypotheses:
x':sort[a33],xs:sort[a34].del'(max(cons(x', xs)), cons(x', xs))=true
x':sort[a33],y:sort[a33].equal_sort[a46](ge(x', y), true_renamed)=true
x':sort[a33],xs:sort[a34].equal_sort[a46](eq(max(cons(x', xs)), x'), true_renamed)=false






----------------------------------------

(45)
Complex Obligation (AND)

----------------------------------------

(46)
Obligation:
Formula:
true=true

Hypotheses:
x':sort[a33],xs:sort[a34].del'(max(cons(x', xs)), cons(x', xs))=true
x':sort[a33],y:sort[a33].equal_sort[a46](ge(x', y), true_renamed)=true
x':sort[a33],xs:sort[a34].equal_sort[a46](eq(max(cons(x', xs)), x'), true_renamed)=true




----------------------------------------

(47) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(48)
YES

----------------------------------------

(49)
Obligation:
Formula:
x':sort[a33],xs:sort[a34],y:sort[a33].del'(max(cons(x', xs)), cons(y, xs))=true

Hypotheses:
x':sort[a33],xs:sort[a34].del'(max(cons(x', xs)), cons(x', xs))=true
x':sort[a33],y:sort[a33].equal_sort[a46](ge(x', y), true_renamed)=true
x':sort[a33],xs:sort[a34].equal_sort[a46](eq(max(cons(x', xs)), x'), true_renamed)=false




----------------------------------------

(50) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
true=true

Hypotheses:
x':sort[a33],xs:sort[a34].del'(max(cons(x', xs)), cons(x', xs))=true
x':sort[a33],y:sort[a33].equal_sort[a46](ge(x', y), true_renamed)=true
x':sort[a33],xs:sort[a34].equal_sort[a46](eq(max(cons(x', xs)), x'), true_renamed)=false
x':sort[a33],xs:sort[a34],y:sort[a33].equal_sort[a46](eq(max(cons(x', xs)), y), true_renamed)=true





Formula:
x':sort[a33],xs:sort[a34].del'(max(cons(x', xs)), xs)=true

Hypotheses:
x':sort[a33],xs:sort[a34].del'(max(cons(x', xs)), cons(x', xs))=true
x':sort[a33],y:sort[a33].equal_sort[a46](ge(x', y), true_renamed)=true
x':sort[a33],xs:sort[a34].equal_sort[a46](eq(max(cons(x', xs)), x'), true_renamed)=false
x':sort[a33],xs:sort[a34],y:sort[a33].equal_sort[a46](eq(max(cons(x', xs)), y), true_renamed)=false






----------------------------------------

(51)
Complex Obligation (AND)

----------------------------------------

(52)
Obligation:
Formula:
true=true

Hypotheses:
x':sort[a33],xs:sort[a34].del'(max(cons(x', xs)), cons(x', xs))=true
x':sort[a33],y:sort[a33].equal_sort[a46](ge(x', y), true_renamed)=true
x':sort[a33],xs:sort[a34].equal_sort[a46](eq(max(cons(x', xs)), x'), true_renamed)=false
x':sort[a33],xs:sort[a34],y:sort[a33].equal_sort[a46](eq(max(cons(x', xs)), y), true_renamed)=true




----------------------------------------

(53) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(54)
YES

----------------------------------------

(55)
Obligation:
Formula:
x':sort[a33],xs:sort[a34].del'(max(cons(x', xs)), xs)=true

Hypotheses:
x':sort[a33],xs:sort[a34].del'(max(cons(x', xs)), cons(x', xs))=true
x':sort[a33],y:sort[a33].equal_sort[a46](ge(x', y), true_renamed)=true
x':sort[a33],xs:sort[a34].equal_sort[a46](eq(max(cons(x', xs)), x'), true_renamed)=false
x':sort[a33],xs:sort[a34],y:sort[a33].equal_sort[a46](eq(max(cons(x', xs)), y), true_renamed)=false




----------------------------------------

(56) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
x':sort[a33],xs:sort[a34].(del'(max(cons(x', xs)), cons(x', xs))=true->del'(max(cons(x', xs)), xs)=true)

Hypotheses:
x':sort[a33],y:sort[a33].equal_sort[a46](ge(x', y), true_renamed)=true
x':sort[a33],xs:sort[a34].equal_sort[a46](eq(max(cons(x', xs)), x'), true_renamed)=false
x':sort[a33],xs:sort[a34],y:sort[a33].equal_sort[a46](eq(max(cons(x', xs)), y), true_renamed)=false




----------------------------------------

(57)
Obligation:
Formula:
x':sort[a33],xs:sort[a34].(del'(max(cons(x', xs)), cons(x', xs))=true->del'(max(cons(x', xs)), xs)=true)

Hypotheses:
x':sort[a33],y:sort[a33].equal_sort[a46](ge(x', y), true_renamed)=true
x':sort[a33],xs:sort[a34].equal_sort[a46](eq(max(cons(x', xs)), x'), true_renamed)=false
x':sort[a33],xs:sort[a34],y:sort[a33].equal_sort[a46](eq(max(cons(x', xs)), y), true_renamed)=false




----------------------------------------

(58) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
x':sort[a33],xs:sort[a34].(del'(max(cons(x', xs)), xs)=true->del'(max(cons(x', xs)), xs)=true)

Hypotheses:
x':sort[a33],y:sort[a33].equal_sort[a46](ge(x', y), true_renamed)=true
x':sort[a33],xs:sort[a34].equal_sort[a46](eq(max(cons(x', xs)), x'), true_renamed)=false
x':sort[a33],xs:sort[a34],y:sort[a33].equal_sort[a46](eq(max(cons(x', xs)), y), true_renamed)=false






----------------------------------------

(59)
Obligation:
Formula:
x':sort[a33],xs:sort[a34].(del'(max(cons(x', xs)), xs)=true->del'(max(cons(x', xs)), xs)=true)

Hypotheses:
x':sort[a33],y:sort[a33].equal_sort[a46](ge(x', y), true_renamed)=true
x':sort[a33],xs:sort[a34].equal_sort[a46](eq(max(cons(x', xs)), x'), true_renamed)=false
x':sort[a33],xs:sort[a34],y:sort[a33].equal_sort[a46](eq(max(cons(x', xs)), y), true_renamed)=false




----------------------------------------

(60) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(61)
YES

----------------------------------------

(62)
Obligation:
Formula:
x':sort[a33],y:sort[a33],xs:sort[a34].(~(cons(x', cons(y, xs))=nil)->del'(max(cons(x', cons(y, xs))), cons(x', cons(y, xs)))=true)

Hypotheses:
y:sort[a33],xs:sort[a34].del'(max(cons(y, xs)), cons(y, xs))=true
x':sort[a33],y:sort[a33].equal_sort[a46](ge(x', y), true_renamed)=false




----------------------------------------

(63) Symbolic evaluation (EQUIVALENT)
Could be shown by simple symbolic evaluation.
----------------------------------------

(64)
Obligation:
Formula:
x':sort[a33],y:sort[a33],xs:sort[a34].del'(max(cons(x', cons(y, xs))), cons(x', cons(y, xs)))=true

Hypotheses:
y:sort[a33],xs:sort[a34].del'(max(cons(y, xs)), cons(y, xs))=true
x':sort[a33],y:sort[a33].equal_sort[a46](ge(x', y), true_renamed)=false




----------------------------------------

(65) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
y:sort[a33],xs:sort[a34],x':sort[a33].del'(max(cons(y, xs)), cons(x', cons(y, xs)))=true

Hypotheses:
y:sort[a33],xs:sort[a34].del'(max(cons(y, xs)), cons(y, xs))=true
x':sort[a33],y:sort[a33].equal_sort[a46](ge(x', y), true_renamed)=false






----------------------------------------

(66)
Obligation:
Formula:
y:sort[a33],xs:sort[a34],x':sort[a33].del'(max(cons(y, xs)), cons(x', cons(y, xs)))=true

Hypotheses:
y:sort[a33],xs:sort[a34].del'(max(cons(y, xs)), cons(y, xs))=true
x':sort[a33],y:sort[a33].equal_sort[a46](ge(x', y), true_renamed)=false




----------------------------------------

(67) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
true=true

Hypotheses:
y:sort[a33],xs:sort[a34].del'(max(cons(y, xs)), cons(y, xs))=true
x':sort[a33],y:sort[a33].equal_sort[a46](ge(x', y), true_renamed)=false
y:sort[a33],xs:sort[a34],x':sort[a33].equal_sort[a46](eq(max(cons(y, xs)), x'), true_renamed)=true





Formula:
y:sort[a33],xs:sort[a34].del'(max(cons(y, xs)), cons(y, xs))=true

Hypotheses:
y:sort[a33],xs:sort[a34].del'(max(cons(y, xs)), cons(y, xs))=true
x':sort[a33],y:sort[a33].equal_sort[a46](ge(x', y), true_renamed)=false
y:sort[a33],xs:sort[a34],x':sort[a33].equal_sort[a46](eq(max(cons(y, xs)), x'), true_renamed)=false






----------------------------------------

(68)
Complex Obligation (AND)

----------------------------------------

(69)
Obligation:
Formula:
true=true

Hypotheses:
y:sort[a33],xs:sort[a34].del'(max(cons(y, xs)), cons(y, xs))=true
x':sort[a33],y:sort[a33].equal_sort[a46](ge(x', y), true_renamed)=false
y:sort[a33],xs:sort[a34],x':sort[a33].equal_sort[a46](eq(max(cons(y, xs)), x'), true_renamed)=true




----------------------------------------

(70) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(71)
YES

----------------------------------------

(72)
Obligation:
Formula:
y:sort[a33],xs:sort[a34].del'(max(cons(y, xs)), cons(y, xs))=true

Hypotheses:
y:sort[a33],xs:sort[a34].del'(max(cons(y, xs)), cons(y, xs))=true
x':sort[a33],y:sort[a33].equal_sort[a46](ge(x', y), true_renamed)=false
y:sort[a33],xs:sort[a34],x':sort[a33].equal_sort[a46](eq(max(cons(y, xs)), x'), true_renamed)=false




----------------------------------------

(73) Symbolic evaluation under hypothesis (EQUIVALENT)
Could be shown using symbolic evaluation under hypothesis, by using the following hypotheses:

y:sort[a33],xs:sort[a34].del'(max(cons(y, xs)), cons(y, xs))=true

----------------------------------------

(74)
YES

(53) Complex Obligation (AND)

(54) Obligation:

Q DP problem:
The TRS P consists of the following rules:

SORT(cons(x0, x1)) → IF3(false, cons(x0, x1))

The TRS R consists of the following rules:

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(55) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(56) TRUE

(57) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

del'(x3, nil) → false
del'(x4, cons(y1, xs1)) → if2'(eq(x4, y1), x4, y1, xs1)
if2'(true_renamed, x7, y4, xs2) → true
if2'(false_renamed, x8, y5, xs3) → del'(x8, xs3)
max(nil) → 0
max(cons(x, nil)) → x
max(cons(x', cons(y, xs))) → if1(ge(x', y), x', y, xs)
if1(true_renamed, x'', y', xs') → max(cons(x'', xs'))
if1(false_renamed, x2, y'', xs'') → max(cons(y'', xs''))
del(x3, nil) → nil
del(x4, cons(y1, xs1)) → if2(eq(x4, y1), x4, y1, xs1)
eq(0, 0) → true_renamed
eq(0, s(y2)) → false_renamed
eq(s(x5), 0) → false_renamed
eq(s(x6), s(y3)) → eq(x6, y3)
if2(true_renamed, x7, y4, xs2) → xs2
if2(false_renamed, x8, y5, xs3) → cons(y5, del(x8, xs3))
ge(x9, 0) → true_renamed
ge(0, s(x10)) → false_renamed
ge(s(x11), s(y6)) → ge(x11, y6)
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a33](0, 0) → true
equal_sort[a33](0, s(v54)) → false
equal_sort[a33](s(v55), 0) → false
equal_sort[a33](s(v55), s(v56)) → equal_sort[a33](v55, v56)
equal_sort[a34](nil, nil) → true
equal_sort[a34](nil, cons(v57, v58)) → false
equal_sort[a34](cons(v59, v60), nil) → false
equal_sort[a34](cons(v59, v60), cons(v61, v62)) → and(equal_sort[a33](v59, v61), equal_sort[a34](v60, v62))
equal_sort[a46](true_renamed, true_renamed) → true
equal_sort[a46](true_renamed, false_renamed) → false
equal_sort[a46](false_renamed, true_renamed) → false
equal_sort[a46](false_renamed, false_renamed) → true
equal_sort[a63](witness_sort[a63], witness_sort[a63]) → true

Q is empty.

(58) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(59) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

del'(x3, nil) → false
del'(x4, cons(y1, xs1)) → if2'(eq(x4, y1), x4, y1, xs1)
if2'(true_renamed, x7, y4, xs2) → true
if2'(false_renamed, x8, y5, xs3) → del'(x8, xs3)
max(nil) → 0
max(cons(x, nil)) → x
max(cons(x', cons(y, xs))) → if1(ge(x', y), x', y, xs)
if1(true_renamed, x'', y', xs') → max(cons(x'', xs'))
if1(false_renamed, x2, y'', xs'') → max(cons(y'', xs''))
del(x3, nil) → nil
del(x4, cons(y1, xs1)) → if2(eq(x4, y1), x4, y1, xs1)
eq(0, 0) → true_renamed
eq(0, s(y2)) → false_renamed
eq(s(x5), 0) → false_renamed
eq(s(x6), s(y3)) → eq(x6, y3)
if2(true_renamed, x7, y4, xs2) → xs2
if2(false_renamed, x8, y5, xs3) → cons(y5, del(x8, xs3))
ge(x9, 0) → true_renamed
ge(0, s(x10)) → false_renamed
ge(s(x11), s(y6)) → ge(x11, y6)
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a33](0, 0) → true
equal_sort[a33](0, s(v54)) → false
equal_sort[a33](s(v55), 0) → false
equal_sort[a33](s(v55), s(v56)) → equal_sort[a33](v55, v56)
equal_sort[a34](nil, nil) → true
equal_sort[a34](nil, cons(v57, v58)) → false
equal_sort[a34](cons(v59, v60), nil) → false
equal_sort[a34](cons(v59, v60), cons(v61, v62)) → and(equal_sort[a33](v59, v61), equal_sort[a34](v60, v62))
equal_sort[a46](true_renamed, true_renamed) → true
equal_sort[a46](true_renamed, false_renamed) → false
equal_sort[a46](false_renamed, true_renamed) → false
equal_sort[a46](false_renamed, false_renamed) → true
equal_sort[a63](witness_sort[a63], witness_sort[a63]) → true

The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true_renamed, x0, x1, x2)
if2(false_renamed, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a46](true_renamed, true_renamed)
equal_sort[a46](true_renamed, false_renamed)
equal_sort[a46](false_renamed, true_renamed)
equal_sort[a46](false_renamed, false_renamed)
equal_sort[a63](witness_sort[a63], witness_sort[a63])

(60) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(61) Obligation:

Q DP problem:
The TRS P consists of the following rules:

DEL'(x4, cons(y1, xs1)) → IF2'(eq(x4, y1), x4, y1, xs1)
DEL'(x4, cons(y1, xs1)) → EQ(x4, y1)
IF2'(false_renamed, x8, y5, xs3) → DEL'(x8, xs3)
MAX(cons(x', cons(y, xs))) → IF1(ge(x', y), x', y, xs)
MAX(cons(x', cons(y, xs))) → GE(x', y)
IF1(true_renamed, x'', y', xs') → MAX(cons(x'', xs'))
IF1(false_renamed, x2, y'', xs'') → MAX(cons(y'', xs''))
DEL(x4, cons(y1, xs1)) → IF2(eq(x4, y1), x4, y1, xs1)
DEL(x4, cons(y1, xs1)) → EQ(x4, y1)
EQ(s(x6), s(y3)) → EQ(x6, y3)
IF2(false_renamed, x8, y5, xs3) → DEL(x8, xs3)
GE(s(x11), s(y6)) → GE(x11, y6)
EQUAL_SORT[A33](s(v55), s(v56)) → EQUAL_SORT[A33](v55, v56)
EQUAL_SORT[A34](cons(v59, v60), cons(v61, v62)) → AND(equal_sort[a33](v59, v61), equal_sort[a34](v60, v62))
EQUAL_SORT[A34](cons(v59, v60), cons(v61, v62)) → EQUAL_SORT[A33](v59, v61)
EQUAL_SORT[A34](cons(v59, v60), cons(v61, v62)) → EQUAL_SORT[A34](v60, v62)

The TRS R consists of the following rules:

del'(x3, nil) → false
del'(x4, cons(y1, xs1)) → if2'(eq(x4, y1), x4, y1, xs1)
if2'(true_renamed, x7, y4, xs2) → true
if2'(false_renamed, x8, y5, xs3) → del'(x8, xs3)
max(nil) → 0
max(cons(x, nil)) → x
max(cons(x', cons(y, xs))) → if1(ge(x', y), x', y, xs)
if1(true_renamed, x'', y', xs') → max(cons(x'', xs'))
if1(false_renamed, x2, y'', xs'') → max(cons(y'', xs''))
del(x3, nil) → nil
del(x4, cons(y1, xs1)) → if2(eq(x4, y1), x4, y1, xs1)
eq(0, 0) → true_renamed
eq(0, s(y2)) → false_renamed
eq(s(x5), 0) → false_renamed
eq(s(x6), s(y3)) → eq(x6, y3)
if2(true_renamed, x7, y4, xs2) → xs2
if2(false_renamed, x8, y5, xs3) → cons(y5, del(x8, xs3))
ge(x9, 0) → true_renamed
ge(0, s(x10)) → false_renamed
ge(s(x11), s(y6)) → ge(x11, y6)
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a33](0, 0) → true
equal_sort[a33](0, s(v54)) → false
equal_sort[a33](s(v55), 0) → false
equal_sort[a33](s(v55), s(v56)) → equal_sort[a33](v55, v56)
equal_sort[a34](nil, nil) → true
equal_sort[a34](nil, cons(v57, v58)) → false
equal_sort[a34](cons(v59, v60), nil) → false
equal_sort[a34](cons(v59, v60), cons(v61, v62)) → and(equal_sort[a33](v59, v61), equal_sort[a34](v60, v62))
equal_sort[a46](true_renamed, true_renamed) → true
equal_sort[a46](true_renamed, false_renamed) → false
equal_sort[a46](false_renamed, true_renamed) → false
equal_sort[a46](false_renamed, false_renamed) → true
equal_sort[a63](witness_sort[a63], witness_sort[a63]) → true

The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true_renamed, x0, x1, x2)
if2(false_renamed, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a46](true_renamed, true_renamed)
equal_sort[a46](true_renamed, false_renamed)
equal_sort[a46](false_renamed, true_renamed)
equal_sort[a46](false_renamed, false_renamed)
equal_sort[a63](witness_sort[a63], witness_sort[a63])

We have to consider all minimal (P,Q,R)-chains.

(62) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 7 SCCs with 5 less nodes.

(63) Complex Obligation (AND)

(64) Obligation:

Q DP problem:
The TRS P consists of the following rules:

EQUAL_SORT[A33](s(v55), s(v56)) → EQUAL_SORT[A33](v55, v56)

The TRS R consists of the following rules:

del'(x3, nil) → false
del'(x4, cons(y1, xs1)) → if2'(eq(x4, y1), x4, y1, xs1)
if2'(true_renamed, x7, y4, xs2) → true
if2'(false_renamed, x8, y5, xs3) → del'(x8, xs3)
max(nil) → 0
max(cons(x, nil)) → x
max(cons(x', cons(y, xs))) → if1(ge(x', y), x', y, xs)
if1(true_renamed, x'', y', xs') → max(cons(x'', xs'))
if1(false_renamed, x2, y'', xs'') → max(cons(y'', xs''))
del(x3, nil) → nil
del(x4, cons(y1, xs1)) → if2(eq(x4, y1), x4, y1, xs1)
eq(0, 0) → true_renamed
eq(0, s(y2)) → false_renamed
eq(s(x5), 0) → false_renamed
eq(s(x6), s(y3)) → eq(x6, y3)
if2(true_renamed, x7, y4, xs2) → xs2
if2(false_renamed, x8, y5, xs3) → cons(y5, del(x8, xs3))
ge(x9, 0) → true_renamed
ge(0, s(x10)) → false_renamed
ge(s(x11), s(y6)) → ge(x11, y6)
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a33](0, 0) → true
equal_sort[a33](0, s(v54)) → false
equal_sort[a33](s(v55), 0) → false
equal_sort[a33](s(v55), s(v56)) → equal_sort[a33](v55, v56)
equal_sort[a34](nil, nil) → true
equal_sort[a34](nil, cons(v57, v58)) → false
equal_sort[a34](cons(v59, v60), nil) → false
equal_sort[a34](cons(v59, v60), cons(v61, v62)) → and(equal_sort[a33](v59, v61), equal_sort[a34](v60, v62))
equal_sort[a46](true_renamed, true_renamed) → true
equal_sort[a46](true_renamed, false_renamed) → false
equal_sort[a46](false_renamed, true_renamed) → false
equal_sort[a46](false_renamed, false_renamed) → true
equal_sort[a63](witness_sort[a63], witness_sort[a63]) → true

The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true_renamed, x0, x1, x2)
if2(false_renamed, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a46](true_renamed, true_renamed)
equal_sort[a46](true_renamed, false_renamed)
equal_sort[a46](false_renamed, true_renamed)
equal_sort[a46](false_renamed, false_renamed)
equal_sort[a63](witness_sort[a63], witness_sort[a63])

We have to consider all minimal (P,Q,R)-chains.

(65) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(66) Obligation:

Q DP problem:
The TRS P consists of the following rules:

EQUAL_SORT[A33](s(v55), s(v56)) → EQUAL_SORT[A33](v55, v56)

R is empty.
The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true_renamed, x0, x1, x2)
if2(false_renamed, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a46](true_renamed, true_renamed)
equal_sort[a46](true_renamed, false_renamed)
equal_sort[a46](false_renamed, true_renamed)
equal_sort[a46](false_renamed, false_renamed)
equal_sort[a63](witness_sort[a63], witness_sort[a63])

We have to consider all minimal (P,Q,R)-chains.

(67) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true_renamed, x0, x1, x2)
if2(false_renamed, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a46](true_renamed, true_renamed)
equal_sort[a46](true_renamed, false_renamed)
equal_sort[a46](false_renamed, true_renamed)
equal_sort[a46](false_renamed, false_renamed)
equal_sort[a63](witness_sort[a63], witness_sort[a63])

(68) Obligation:

Q DP problem:
The TRS P consists of the following rules:

EQUAL_SORT[A33](s(v55), s(v56)) → EQUAL_SORT[A33](v55, v56)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(69) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

EQUAL_SORT[A33](s(v55), s(v56)) → EQUAL_SORT[A33](v55, v56)
The graph contains the following edges 1 > 1, 2 > 2

(70) YES

(71) Obligation:

Q DP problem:
The TRS P consists of the following rules:

EQUAL_SORT[A34](cons(v59, v60), cons(v61, v62)) → EQUAL_SORT[A34](v60, v62)

The TRS R consists of the following rules:

del'(x3, nil) → false
del'(x4, cons(y1, xs1)) → if2'(eq(x4, y1), x4, y1, xs1)
if2'(true_renamed, x7, y4, xs2) → true
if2'(false_renamed, x8, y5, xs3) → del'(x8, xs3)
max(nil) → 0
max(cons(x, nil)) → x
max(cons(x', cons(y, xs))) → if1(ge(x', y), x', y, xs)
if1(true_renamed, x'', y', xs') → max(cons(x'', xs'))
if1(false_renamed, x2, y'', xs'') → max(cons(y'', xs''))
del(x3, nil) → nil
del(x4, cons(y1, xs1)) → if2(eq(x4, y1), x4, y1, xs1)
eq(0, 0) → true_renamed
eq(0, s(y2)) → false_renamed
eq(s(x5), 0) → false_renamed
eq(s(x6), s(y3)) → eq(x6, y3)
if2(true_renamed, x7, y4, xs2) → xs2
if2(false_renamed, x8, y5, xs3) → cons(y5, del(x8, xs3))
ge(x9, 0) → true_renamed
ge(0, s(x10)) → false_renamed
ge(s(x11), s(y6)) → ge(x11, y6)
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a33](0, 0) → true
equal_sort[a33](0, s(v54)) → false
equal_sort[a33](s(v55), 0) → false
equal_sort[a33](s(v55), s(v56)) → equal_sort[a33](v55, v56)
equal_sort[a34](nil, nil) → true
equal_sort[a34](nil, cons(v57, v58)) → false
equal_sort[a34](cons(v59, v60), nil) → false
equal_sort[a34](cons(v59, v60), cons(v61, v62)) → and(equal_sort[a33](v59, v61), equal_sort[a34](v60, v62))
equal_sort[a46](true_renamed, true_renamed) → true
equal_sort[a46](true_renamed, false_renamed) → false
equal_sort[a46](false_renamed, true_renamed) → false
equal_sort[a46](false_renamed, false_renamed) → true
equal_sort[a63](witness_sort[a63], witness_sort[a63]) → true

The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true_renamed, x0, x1, x2)
if2(false_renamed, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a46](true_renamed, true_renamed)
equal_sort[a46](true_renamed, false_renamed)
equal_sort[a46](false_renamed, true_renamed)
equal_sort[a46](false_renamed, false_renamed)
equal_sort[a63](witness_sort[a63], witness_sort[a63])

We have to consider all minimal (P,Q,R)-chains.

(72) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(73) Obligation:

Q DP problem:
The TRS P consists of the following rules:

EQUAL_SORT[A34](cons(v59, v60), cons(v61, v62)) → EQUAL_SORT[A34](v60, v62)

R is empty.
The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true_renamed, x0, x1, x2)
if2(false_renamed, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a46](true_renamed, true_renamed)
equal_sort[a46](true_renamed, false_renamed)
equal_sort[a46](false_renamed, true_renamed)
equal_sort[a46](false_renamed, false_renamed)
equal_sort[a63](witness_sort[a63], witness_sort[a63])

We have to consider all minimal (P,Q,R)-chains.

(74) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true_renamed, x0, x1, x2)
if2(false_renamed, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a46](true_renamed, true_renamed)
equal_sort[a46](true_renamed, false_renamed)
equal_sort[a46](false_renamed, true_renamed)
equal_sort[a46](false_renamed, false_renamed)
equal_sort[a63](witness_sort[a63], witness_sort[a63])

(75) Obligation:

Q DP problem:
The TRS P consists of the following rules:

EQUAL_SORT[A34](cons(v59, v60), cons(v61, v62)) → EQUAL_SORT[A34](v60, v62)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(76) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

EQUAL_SORT[A34](cons(v59, v60), cons(v61, v62)) → EQUAL_SORT[A34](v60, v62)
The graph contains the following edges 1 > 1, 2 > 2

(77) YES

(78) Obligation:

Q DP problem:
The TRS P consists of the following rules:

GE(s(x11), s(y6)) → GE(x11, y6)

The TRS R consists of the following rules:

del'(x3, nil) → false
del'(x4, cons(y1, xs1)) → if2'(eq(x4, y1), x4, y1, xs1)
if2'(true_renamed, x7, y4, xs2) → true
if2'(false_renamed, x8, y5, xs3) → del'(x8, xs3)
max(nil) → 0
max(cons(x, nil)) → x
max(cons(x', cons(y, xs))) → if1(ge(x', y), x', y, xs)
if1(true_renamed, x'', y', xs') → max(cons(x'', xs'))
if1(false_renamed, x2, y'', xs'') → max(cons(y'', xs''))
del(x3, nil) → nil
del(x4, cons(y1, xs1)) → if2(eq(x4, y1), x4, y1, xs1)
eq(0, 0) → true_renamed
eq(0, s(y2)) → false_renamed
eq(s(x5), 0) → false_renamed
eq(s(x6), s(y3)) → eq(x6, y3)
if2(true_renamed, x7, y4, xs2) → xs2
if2(false_renamed, x8, y5, xs3) → cons(y5, del(x8, xs3))
ge(x9, 0) → true_renamed
ge(0, s(x10)) → false_renamed
ge(s(x11), s(y6)) → ge(x11, y6)
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a33](0, 0) → true
equal_sort[a33](0, s(v54)) → false
equal_sort[a33](s(v55), 0) → false
equal_sort[a33](s(v55), s(v56)) → equal_sort[a33](v55, v56)
equal_sort[a34](nil, nil) → true
equal_sort[a34](nil, cons(v57, v58)) → false
equal_sort[a34](cons(v59, v60), nil) → false
equal_sort[a34](cons(v59, v60), cons(v61, v62)) → and(equal_sort[a33](v59, v61), equal_sort[a34](v60, v62))
equal_sort[a46](true_renamed, true_renamed) → true
equal_sort[a46](true_renamed, false_renamed) → false
equal_sort[a46](false_renamed, true_renamed) → false
equal_sort[a46](false_renamed, false_renamed) → true
equal_sort[a63](witness_sort[a63], witness_sort[a63]) → true

The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true_renamed, x0, x1, x2)
if2(false_renamed, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a46](true_renamed, true_renamed)
equal_sort[a46](true_renamed, false_renamed)
equal_sort[a46](false_renamed, true_renamed)
equal_sort[a46](false_renamed, false_renamed)
equal_sort[a63](witness_sort[a63], witness_sort[a63])

We have to consider all minimal (P,Q,R)-chains.

(79) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(80) Obligation:

Q DP problem:
The TRS P consists of the following rules:

GE(s(x11), s(y6)) → GE(x11, y6)

R is empty.
The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true_renamed, x0, x1, x2)
if2(false_renamed, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a46](true_renamed, true_renamed)
equal_sort[a46](true_renamed, false_renamed)
equal_sort[a46](false_renamed, true_renamed)
equal_sort[a46](false_renamed, false_renamed)
equal_sort[a63](witness_sort[a63], witness_sort[a63])

We have to consider all minimal (P,Q,R)-chains.

(81) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true_renamed, x0, x1, x2)
if2(false_renamed, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a46](true_renamed, true_renamed)
equal_sort[a46](true_renamed, false_renamed)
equal_sort[a46](false_renamed, true_renamed)
equal_sort[a46](false_renamed, false_renamed)
equal_sort[a63](witness_sort[a63], witness_sort[a63])

(82) Obligation:

Q DP problem:
The TRS P consists of the following rules:

GE(s(x11), s(y6)) → GE(x11, y6)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(83) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

GE(s(x11), s(y6)) → GE(x11, y6)
The graph contains the following edges 1 > 1, 2 > 2

(84) YES

(85) Obligation:

Q DP problem:
The TRS P consists of the following rules:

EQ(s(x6), s(y3)) → EQ(x6, y3)

The TRS R consists of the following rules:

del'(x3, nil) → false
del'(x4, cons(y1, xs1)) → if2'(eq(x4, y1), x4, y1, xs1)
if2'(true_renamed, x7, y4, xs2) → true
if2'(false_renamed, x8, y5, xs3) → del'(x8, xs3)
max(nil) → 0
max(cons(x, nil)) → x
max(cons(x', cons(y, xs))) → if1(ge(x', y), x', y, xs)
if1(true_renamed, x'', y', xs') → max(cons(x'', xs'))
if1(false_renamed, x2, y'', xs'') → max(cons(y'', xs''))
del(x3, nil) → nil
del(x4, cons(y1, xs1)) → if2(eq(x4, y1), x4, y1, xs1)
eq(0, 0) → true_renamed
eq(0, s(y2)) → false_renamed
eq(s(x5), 0) → false_renamed
eq(s(x6), s(y3)) → eq(x6, y3)
if2(true_renamed, x7, y4, xs2) → xs2
if2(false_renamed, x8, y5, xs3) → cons(y5, del(x8, xs3))
ge(x9, 0) → true_renamed
ge(0, s(x10)) → false_renamed
ge(s(x11), s(y6)) → ge(x11, y6)
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a33](0, 0) → true
equal_sort[a33](0, s(v54)) → false
equal_sort[a33](s(v55), 0) → false
equal_sort[a33](s(v55), s(v56)) → equal_sort[a33](v55, v56)
equal_sort[a34](nil, nil) → true
equal_sort[a34](nil, cons(v57, v58)) → false
equal_sort[a34](cons(v59, v60), nil) → false
equal_sort[a34](cons(v59, v60), cons(v61, v62)) → and(equal_sort[a33](v59, v61), equal_sort[a34](v60, v62))
equal_sort[a46](true_renamed, true_renamed) → true
equal_sort[a46](true_renamed, false_renamed) → false
equal_sort[a46](false_renamed, true_renamed) → false
equal_sort[a46](false_renamed, false_renamed) → true
equal_sort[a63](witness_sort[a63], witness_sort[a63]) → true

The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true_renamed, x0, x1, x2)
if2(false_renamed, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a46](true_renamed, true_renamed)
equal_sort[a46](true_renamed, false_renamed)
equal_sort[a46](false_renamed, true_renamed)
equal_sort[a46](false_renamed, false_renamed)
equal_sort[a63](witness_sort[a63], witness_sort[a63])

We have to consider all minimal (P,Q,R)-chains.

(86) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(87) Obligation:

Q DP problem:
The TRS P consists of the following rules:

EQ(s(x6), s(y3)) → EQ(x6, y3)

R is empty.
The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true_renamed, x0, x1, x2)
if2(false_renamed, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a46](true_renamed, true_renamed)
equal_sort[a46](true_renamed, false_renamed)
equal_sort[a46](false_renamed, true_renamed)
equal_sort[a46](false_renamed, false_renamed)
equal_sort[a63](witness_sort[a63], witness_sort[a63])

We have to consider all minimal (P,Q,R)-chains.

(88) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true_renamed, x0, x1, x2)
if2(false_renamed, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a46](true_renamed, true_renamed)
equal_sort[a46](true_renamed, false_renamed)
equal_sort[a46](false_renamed, true_renamed)
equal_sort[a46](false_renamed, false_renamed)
equal_sort[a63](witness_sort[a63], witness_sort[a63])

(89) Obligation:

Q DP problem:
The TRS P consists of the following rules:

EQ(s(x6), s(y3)) → EQ(x6, y3)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(90) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

EQ(s(x6), s(y3)) → EQ(x6, y3)
The graph contains the following edges 1 > 1, 2 > 2

(91) YES

(92) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF2(false_renamed, x8, y5, xs3) → DEL(x8, xs3)
DEL(x4, cons(y1, xs1)) → IF2(eq(x4, y1), x4, y1, xs1)

The TRS R consists of the following rules:

del'(x3, nil) → false
del'(x4, cons(y1, xs1)) → if2'(eq(x4, y1), x4, y1, xs1)
if2'(true_renamed, x7, y4, xs2) → true
if2'(false_renamed, x8, y5, xs3) → del'(x8, xs3)
max(nil) → 0
max(cons(x, nil)) → x
max(cons(x', cons(y, xs))) → if1(ge(x', y), x', y, xs)
if1(true_renamed, x'', y', xs') → max(cons(x'', xs'))
if1(false_renamed, x2, y'', xs'') → max(cons(y'', xs''))
del(x3, nil) → nil
del(x4, cons(y1, xs1)) → if2(eq(x4, y1), x4, y1, xs1)
eq(0, 0) → true_renamed
eq(0, s(y2)) → false_renamed
eq(s(x5), 0) → false_renamed
eq(s(x6), s(y3)) → eq(x6, y3)
if2(true_renamed, x7, y4, xs2) → xs2
if2(false_renamed, x8, y5, xs3) → cons(y5, del(x8, xs3))
ge(x9, 0) → true_renamed
ge(0, s(x10)) → false_renamed
ge(s(x11), s(y6)) → ge(x11, y6)
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a33](0, 0) → true
equal_sort[a33](0, s(v54)) → false
equal_sort[a33](s(v55), 0) → false
equal_sort[a33](s(v55), s(v56)) → equal_sort[a33](v55, v56)
equal_sort[a34](nil, nil) → true
equal_sort[a34](nil, cons(v57, v58)) → false
equal_sort[a34](cons(v59, v60), nil) → false
equal_sort[a34](cons(v59, v60), cons(v61, v62)) → and(equal_sort[a33](v59, v61), equal_sort[a34](v60, v62))
equal_sort[a46](true_renamed, true_renamed) → true
equal_sort[a46](true_renamed, false_renamed) → false
equal_sort[a46](false_renamed, true_renamed) → false
equal_sort[a46](false_renamed, false_renamed) → true
equal_sort[a63](witness_sort[a63], witness_sort[a63]) → true

The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true_renamed, x0, x1, x2)
if2(false_renamed, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a46](true_renamed, true_renamed)
equal_sort[a46](true_renamed, false_renamed)
equal_sort[a46](false_renamed, true_renamed)
equal_sort[a46](false_renamed, false_renamed)
equal_sort[a63](witness_sort[a63], witness_sort[a63])

We have to consider all minimal (P,Q,R)-chains.

(93) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(94) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF2(false_renamed, x8, y5, xs3) → DEL(x8, xs3)
DEL(x4, cons(y1, xs1)) → IF2(eq(x4, y1), x4, y1, xs1)

The TRS R consists of the following rules:

eq(0, 0) → true_renamed
eq(0, s(y2)) → false_renamed
eq(s(x5), 0) → false_renamed
eq(s(x6), s(y3)) → eq(x6, y3)

The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true_renamed, x0, x1, x2)
if2(false_renamed, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a46](true_renamed, true_renamed)
equal_sort[a46](true_renamed, false_renamed)
equal_sort[a46](false_renamed, true_renamed)
equal_sort[a46](false_renamed, false_renamed)
equal_sort[a63](witness_sort[a63], witness_sort[a63])

We have to consider all minimal (P,Q,R)-chains.

(95) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true_renamed, x0, x1, x2)
if2(false_renamed, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a46](true_renamed, true_renamed)
equal_sort[a46](true_renamed, false_renamed)
equal_sort[a46](false_renamed, true_renamed)
equal_sort[a46](false_renamed, false_renamed)
equal_sort[a63](witness_sort[a63], witness_sort[a63])

(96) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF2(false_renamed, x8, y5, xs3) → DEL(x8, xs3)
DEL(x4, cons(y1, xs1)) → IF2(eq(x4, y1), x4, y1, xs1)

The TRS R consists of the following rules:

eq(0, 0) → true_renamed
eq(0, s(y2)) → false_renamed
eq(s(x5), 0) → false_renamed
eq(s(x6), s(y3)) → eq(x6, y3)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(97) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

DEL(x4, cons(y1, xs1)) → IF2(eq(x4, y1), x4, y1, xs1)
The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4
IF2(false_renamed, x8, y5, xs3) → DEL(x8, xs3)
The graph contains the following edges 2 >= 1, 4 >= 2

(98) YES

(99) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF1(true_renamed, x'', y', xs') → MAX(cons(x'', xs'))
MAX(cons(x', cons(y, xs))) → IF1(ge(x', y), x', y, xs)
IF1(false_renamed, x2, y'', xs'') → MAX(cons(y'', xs''))

The TRS R consists of the following rules:

del'(x3, nil) → false
del'(x4, cons(y1, xs1)) → if2'(eq(x4, y1), x4, y1, xs1)
if2'(true_renamed, x7, y4, xs2) → true
if2'(false_renamed, x8, y5, xs3) → del'(x8, xs3)
max(nil) → 0
max(cons(x, nil)) → x
max(cons(x', cons(y, xs))) → if1(ge(x', y), x', y, xs)
if1(true_renamed, x'', y', xs') → max(cons(x'', xs'))
if1(false_renamed, x2, y'', xs'') → max(cons(y'', xs''))
del(x3, nil) → nil
del(x4, cons(y1, xs1)) → if2(eq(x4, y1), x4, y1, xs1)
eq(0, 0) → true_renamed
eq(0, s(y2)) → false_renamed
eq(s(x5), 0) → false_renamed
eq(s(x6), s(y3)) → eq(x6, y3)
if2(true_renamed, x7, y4, xs2) → xs2
if2(false_renamed, x8, y5, xs3) → cons(y5, del(x8, xs3))
ge(x9, 0) → true_renamed
ge(0, s(x10)) → false_renamed
ge(s(x11), s(y6)) → ge(x11, y6)
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a33](0, 0) → true
equal_sort[a33](0, s(v54)) → false
equal_sort[a33](s(v55), 0) → false
equal_sort[a33](s(v55), s(v56)) → equal_sort[a33](v55, v56)
equal_sort[a34](nil, nil) → true
equal_sort[a34](nil, cons(v57, v58)) → false
equal_sort[a34](cons(v59, v60), nil) → false
equal_sort[a34](cons(v59, v60), cons(v61, v62)) → and(equal_sort[a33](v59, v61), equal_sort[a34](v60, v62))
equal_sort[a46](true_renamed, true_renamed) → true
equal_sort[a46](true_renamed, false_renamed) → false
equal_sort[a46](false_renamed, true_renamed) → false
equal_sort[a46](false_renamed, false_renamed) → true
equal_sort[a63](witness_sort[a63], witness_sort[a63]) → true

The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true_renamed, x0, x1, x2)
if2(false_renamed, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a46](true_renamed, true_renamed)
equal_sort[a46](true_renamed, false_renamed)
equal_sort[a46](false_renamed, true_renamed)
equal_sort[a46](false_renamed, false_renamed)
equal_sort[a63](witness_sort[a63], witness_sort[a63])

We have to consider all minimal (P,Q,R)-chains.

(100) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(101) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF1(true_renamed, x'', y', xs') → MAX(cons(x'', xs'))
MAX(cons(x', cons(y, xs))) → IF1(ge(x', y), x', y, xs)
IF1(false_renamed, x2, y'', xs'') → MAX(cons(y'', xs''))

The TRS R consists of the following rules:

ge(x9, 0) → true_renamed
ge(0, s(x10)) → false_renamed
ge(s(x11), s(y6)) → ge(x11, y6)

The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true_renamed, x0, x1, x2)
if2(false_renamed, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a46](true_renamed, true_renamed)
equal_sort[a46](true_renamed, false_renamed)
equal_sort[a46](false_renamed, true_renamed)
equal_sort[a46](false_renamed, false_renamed)
equal_sort[a63](witness_sort[a63], witness_sort[a63])

We have to consider all minimal (P,Q,R)-chains.

(102) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true_renamed, x0, x1, x2)
if2(false_renamed, x0, x1, x2)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a46](true_renamed, true_renamed)
equal_sort[a46](true_renamed, false_renamed)
equal_sort[a46](false_renamed, true_renamed)
equal_sort[a46](false_renamed, false_renamed)
equal_sort[a63](witness_sort[a63], witness_sort[a63])

(103) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF1(true_renamed, x'', y', xs') → MAX(cons(x'', xs'))
MAX(cons(x', cons(y, xs))) → IF1(ge(x', y), x', y, xs)
IF1(false_renamed, x2, y'', xs'') → MAX(cons(y'', xs''))

The TRS R consists of the following rules:

ge(x9, 0) → true_renamed
ge(0, s(x10)) → false_renamed
ge(s(x11), s(y6)) → ge(x11, y6)

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(104) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

IF1(true_renamed, x'', y', xs') → MAX(cons(x'', xs'))
MAX(cons(x', cons(y, xs))) → IF1(ge(x', y), x', y, xs)
IF1(false_renamed, x2, y'', xs'') → MAX(cons(y'', xs''))

trivial

and weight map:

dummyConstant=1
IF1_1=3
cons_1=2

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none

(105) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

ge(x9, 0) → true_renamed
ge(0, s(x10)) → false_renamed
ge(s(x11), s(y6)) → ge(x11, y6)

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(106) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(107) YES

(108) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF2'(false_renamed, x8, y5, xs3) → DEL'(x8, xs3)
DEL'(x4, cons(y1, xs1)) → IF2'(eq(x4, y1), x4, y1, xs1)

The TRS R consists of the following rules:

del'(x3, nil) → false
del'(x4, cons(y1, xs1)) → if2'(eq(x4, y1), x4, y1, xs1)
if2'(true_renamed, x7, y4, xs2) → true
if2'(false_renamed, x8, y5, xs3) → del'(x8, xs3)
max(nil) → 0
max(cons(x, nil)) → x
max(cons(x', cons(y, xs))) → if1(ge(x', y), x', y, xs)
if1(true_renamed, x'', y', xs') → max(cons(x'', xs'))
if1(false_renamed, x2, y'', xs'') → max(cons(y'', xs''))
del(x3, nil) → nil
del(x4, cons(y1, xs1)) → if2(eq(x4, y1), x4, y1, xs1)
eq(0, 0) → true_renamed
eq(0, s(y2)) → false_renamed
eq(s(x5), 0) → false_renamed
eq(s(x6), s(y3)) → eq(x6, y3)
if2(true_renamed, x7, y4, xs2) → xs2
if2(false_renamed, x8, y5, xs3) → cons(y5, del(x8, xs3))
ge(x9, 0) → true_renamed
ge(0, s(x10)) → false_renamed
ge(s(x11), s(y6)) → ge(x11, y6)
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a33](0, 0) → true
equal_sort[a33](0, s(v54)) → false
equal_sort[a33](s(v55), 0) → false
equal_sort[a33](s(v55), s(v56)) → equal_sort[a33](v55, v56)
equal_sort[a34](nil, nil) → true
equal_sort[a34](nil, cons(v57, v58)) → false
equal_sort[a34](cons(v59, v60), nil) → false
equal_sort[a34](cons(v59, v60), cons(v61, v62)) → and(equal_sort[a33](v59, v61), equal_sort[a34](v60, v62))
equal_sort[a46](true_renamed, true_renamed) → true
equal_sort[a46](true_renamed, false_renamed) → false
equal_sort[a46](false_renamed, true_renamed) → false
equal_sort[a46](false_renamed, false_renamed) → true
equal_sort[a63](witness_sort[a63], witness_sort[a63]) → true

The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true_renamed, x0, x1, x2)
if2(false_renamed, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a46](true_renamed, true_renamed)
equal_sort[a46](true_renamed, false_renamed)
equal_sort[a46](false_renamed, true_renamed)
equal_sort[a46](false_renamed, false_renamed)
equal_sort[a63](witness_sort[a63], witness_sort[a63])

We have to consider all minimal (P,Q,R)-chains.

(109) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(110) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF2'(false_renamed, x8, y5, xs3) → DEL'(x8, xs3)
DEL'(x4, cons(y1, xs1)) → IF2'(eq(x4, y1), x4, y1, xs1)

The TRS R consists of the following rules:

eq(0, 0) → true_renamed
eq(0, s(y2)) → false_renamed
eq(s(x5), 0) → false_renamed
eq(s(x6), s(y3)) → eq(x6, y3)

The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true_renamed, x0, x1, x2)
if2(false_renamed, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a46](true_renamed, true_renamed)
equal_sort[a46](true_renamed, false_renamed)
equal_sort[a46](false_renamed, true_renamed)
equal_sort[a46](false_renamed, false_renamed)
equal_sort[a63](witness_sort[a63], witness_sort[a63])

We have to consider all minimal (P,Q,R)-chains.

(111) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true_renamed, x0, x1, x2)
if2(false_renamed, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a46](true_renamed, true_renamed)
equal_sort[a46](true_renamed, false_renamed)
equal_sort[a46](false_renamed, true_renamed)
equal_sort[a46](false_renamed, false_renamed)
equal_sort[a63](witness_sort[a63], witness_sort[a63])

(112) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF2'(false_renamed, x8, y5, xs3) → DEL'(x8, xs3)
DEL'(x4, cons(y1, xs1)) → IF2'(eq(x4, y1), x4, y1, xs1)

The TRS R consists of the following rules:

eq(0, 0) → true_renamed
eq(0, s(y2)) → false_renamed
eq(s(x5), 0) → false_renamed
eq(s(x6), s(y3)) → eq(x6, y3)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(113) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

DEL'(x4, cons(y1, xs1)) → IF2'(eq(x4, y1), x4, y1, xs1)
The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4
IF2'(false_renamed, x8, y5, xs3) → DEL'(x8, xs3)
The graph contains the following edges 2 >= 1, 4 >= 2