Termination w.r.t. Q proof of AProVE_08_parting03

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 QDPSizeChangeProof (⇔, 0 ms)

        ↳34 YES

    ↳35 QDP

        ↳36 UsableRulesProof (⇔, 0 ms)

        ↳37 QDP

        ↳38 QReductionProof (⇔, 0 ms)

        ↳39 QDP

        ↳40 Induction-Processor (⇒, 466 ms)

        ↳41 AND

            ↳42 QDP

                ↳43 PisEmptyProof (⇔, 0 ms)

                ↳44 YES

            ↳45 QTRS

                ↳46 Overlay + Local Confluence (⇔, 5 ms)

                ↳47 QTRS

                ↳48 DependencyPairsProof (⇔, 0 ms)

                ↳49 QDP

                ↳50 DependencyGraphProof (⇔, 0 ms)

                ↳51 AND

                    ↳52 QDP

                        ↳53 UsableRulesProof (⇔, 0 ms)

                        ↳54 QDP

                        ↳55 QReductionProof (⇔, 0 ms)

                        ↳56 QDP

                        ↳57 QDPSizeChangeProof (⇔, 0 ms)

                        ↳58 YES

                    ↳59 QDP

                        ↳60 UsableRulesProof (⇔, 0 ms)

                        ↳61 QDP

                        ↳62 QReductionProof (⇔, 0 ms)

                        ↳63 QDP

                        ↳64 QDPSizeChangeProof (⇔, 0 ms)

                        ↳65 YES

                    ↳66 QDP

                        ↳67 UsableRulesProof (⇔, 0 ms)

                        ↳68 QDP

                        ↳69 QReductionProof (⇔, 0 ms)

                        ↳70 QDP

                        ↳71 QDPSizeChangeProof (⇔, 0 ms)

                        ↳72 YES

                    ↳73 QDP

                        ↳74 UsableRulesProof (⇔, 0 ms)

                        ↳75 QDP

                        ↳76 QReductionProof (⇔, 0 ms)

                        ↳77 QDP

                        ↳78 QDPSizeChangeProof (⇔, 0 ms)

                        ↳79 YES

                    ↳80 QDP

                        ↳81 UsableRulesProof (⇔, 0 ms)

                        ↳82 QDP

                        ↳83 QReductionProof (⇔, 0 ms)

                        ↳84 QDP

                        ↳85 QDPSizeChangeProof (⇔, 0 ms)

                        ↳86 YES

                    ↳87 QDP

                        ↳88 UsableRulesProof (⇔, 0 ms)

                        ↳89 QDP

                        ↳90 QReductionProof (⇔, 0 ms)

                        ↳91 QDP

                        ↳92 QDPSizeChangeProof (⇔, 0 ms)

                        ↳93 YES

                    ↳94 QDP

                        ↳95 UsableRulesProof (⇔, 0 ms)

                        ↳96 QDP

                        ↳97 QReductionProof (⇔, 0 ms)

                        ↳98 QDP

                        ↳99 QDPSizeChangeProof (⇔, 0 ms)

                        ↳100 YES

(0) Obligation:

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

(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:

LE(s(x), s(y)) → LE(x, y)
EQ(s(x), s(y)) → EQ(x, y)
IF1(true, x, y, xs) → MIN(x, xs)
IF1(false, x, y, xs) → MIN(y, xs)
IF2(false, x, y, xs) → DEL(x, xs)
MINSORT(cons(x, y)) → MIN(x, y)
MINSORT(cons(x, y)) → MINSORT(del(min(x, y), cons(x, y)))
MINSORT(cons(x, y)) → DEL(min(x, y), cons(x, y))
MIN(x, cons(y, z)) → IF1(le(x, y), x, y, z)
MIN(x, cons(y, z)) → LE(x, y)
DEL(x, cons(y, z)) → IF2(eq(x, y), x, y, z)
DEL(x, cons(y, z)) → EQ(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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 4 less nodes.

(6) Complex Obligation (AND)

(7) 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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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:

EQ(s(x), s(y)) → EQ(x, y)

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

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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].

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

(11) 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.

(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:

EQ(s(x), s(y)) → EQ(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:

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

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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:

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

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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].

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

(18) Obligation:

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

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

(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:

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

(20) YES

(21) Obligation:

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

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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:

LE(s(x), s(y)) → LE(x, y)

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

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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].

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

(25) Obligation:

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

LE(s(x), s(y)) → LE(x, y)

R is empty.
Q is empty.
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:

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

(27) YES

(28) Obligation:

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

MIN(x, cons(y, z)) → IF1(le(x, y), x, y, z)
IF1(true, x, y, xs) → MIN(x, xs)
IF1(false, x, y, xs) → MIN(y, xs)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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:

MIN(x, cons(y, z)) → IF1(le(x, y), x, y, z)
IF1(true, x, y, xs) → MIN(x, xs)
IF1(false, x, y, xs) → MIN(y, xs)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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].

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

(32) Obligation:

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

MIN(x, cons(y, z)) → IF1(le(x, y), x, y, z)
IF1(true, x, y, xs) → MIN(x, xs)
IF1(false, x, y, xs) → MIN(y, xs)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

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

(33) 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:

MIN(x, cons(y, z)) → IF1(le(x, y), x, y, z)
The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4
IF1(true, x, y, xs) → MIN(x, xs)
The graph contains the following edges 2 >= 1, 4 >= 2
IF1(false, x, y, xs) → MIN(y, xs)
The graph contains the following edges 3 >= 1, 4 >= 2

(34) YES

(35) Obligation:

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

MINSORT(cons(x, y)) → MINSORT(del(min(x, y), cons(x, y)))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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

(36) 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.

(37) Obligation:

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

MINSORT(cons(x, y)) → MINSORT(del(min(x, y), cons(x, y)))

The TRS R consists of the following rules:

min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)
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))
del(x, nil) → nil
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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

(38) 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].

minsort(nil)
minsort(cons(x0, x1))

(39) Obligation:

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

MINSORT(cons(x, y)) → MINSORT(del(min(x, y), cons(x, y)))

The TRS R consists of the following rules:

min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)
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))
del(x, nil) → nil
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

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

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

(40) Induction-Processor (SOUND transformation)

This DP could be deleted by the Induction-Processor:
MINSORT(cons(x, y)) → MINSORT(del(min(x, y), cons(x, y)))

This order was computed:
Polynomial interpretation [POLO]:

POL(0) = 0
POL(MINSORT(x₁)) = x₁
POL(cons(x₁, x₂)) = 1 + x₁ + x₂
POL(del(x₁, x₂)) = x₂
POL(eq(x₁, x₂)) = x₁
POL(false_renamed) = 0
POL(if1(x₁, x₂, x₃, x₄)) = x₂ + x₃ + x₄
POL(if2(x₁, x₂, x₃, x₄)) = 1 + x₃ + x₄
POL(le(x₁, x₂)) = 1 + x₂
POL(min(x₁, x₂)) = x₁ + x₂
POL(nil) = 0
POL(s(x₁)) = x₁
POL(true_renamed) = 0

At least one of these decreasing rules is always used after the deleted DP:
if2(true_renamed, x6, y5, xs'') → xs''

The following formula is valid:
x:sort[a0],y:sort[a33].del'(min(x , y ), cons(x , y ))=true

The transformed set:
del'(x3, cons(y2, z')) → if2'(eq(x3, y2), x3, y2, z')
if2'(true_renamed, x6, y5, xs'') → true
if2'(false_renamed, x7, y6, xs1) → del'(x7, xs1)
del'(x8, nil) → false
min(x', nil) → x'
min(x'', cons(y', z)) → if1(le(x'', y'), x'', y', z)
if1(true_renamed, x1, y'', xs) → min(x1, xs)
if1(false_renamed, x2, y1, xs') → min(y1, xs')
del(x3, cons(y2, z')) → if2(eq(x3, y2), x3, y2, z')
eq(0, 0) → true_renamed
eq(0, s(y3)) → false_renamed
eq(s(x4), 0) → false_renamed
eq(s(x5), s(y4)) → eq(x5, y4)
if2(true_renamed, x6, y5, xs'') → xs''
if2(false_renamed, x7, y6, xs1) → cons(y6, del(x7, xs1))
del(x8, nil) → nil
le(0, y7) → true_renamed
le(s(x9), 0) → false_renamed
le(s(x10), s(y8)) → le(x10, y8)
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[a0](0, 0) → true
equal_sort[a0](0, s(v25)) → false
equal_sort[a0](s(v26), 0) → false
equal_sort[a0](s(v26), s(v27)) → equal_sort[a0](v26, v27)
equal_sort[a33](nil, nil) → true
equal_sort[a33](nil, cons(v28, v29)) → false
equal_sort[a33](cons(v30, v31), nil) → false
equal_sort[a33](cons(v30, v31), cons(v32, v33)) → and(equal_sort[a0](v30, v32), equal_sort[a33](v31, v33))
equal_sort[a44](true_renamed, true_renamed) → true
equal_sort[a44](true_renamed, false_renamed) → false
equal_sort[a44](false_renamed, true_renamed) → false
equal_sort[a44](false_renamed, false_renamed) → true
equal_sort[a62](witness_sort[a62], witness_sort[a62]) → 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, v25, v26, v27, v28, v29, v30, v31, v32, v33, x6, y5, xs'', x7, y6, y2, z', x8, x3, x', x'', y', z, y3, x4, x5, y4, y7, x9, x10, y8, x1, y'', x2, y1]
   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[a0](0, 0) -> true
   equal_sort[a0](0, s(v25)) -> false
   equal_sort[a0](s(v26), 0) -> false
   equal_sort[a0](s(v26), s(v27)) -> equal_sort[a0](v26, v27)
   equal_sort[a33](nil, nil) -> true
   equal_sort[a33](nil, cons(v28, v29)) -> false
   equal_sort[a33](cons(v30, v31), nil) -> false
   equal_sort[a33](cons(v30, v31), cons(v32, v33)) -> equal_sort[a0](v30, v32) and equal_sort[a33](v31, v33)
   equal_sort[a44](true_renamed, true_renamed) -> true
   equal_sort[a44](true_renamed, false_renamed) -> false
   equal_sort[a44](false_renamed, true_renamed) -> false
   equal_sort[a44](false_renamed, false_renamed) -> true
   equal_sort[a62](witness_sort[a62], witness_sort[a62]) -> true
   if2'(true_renamed, x6, y5, xs'') -> true
   if2'(false_renamed, x7, y6, cons(y2, z')) -> if2'(eq(x7, y2), x7, y2, z')
   if2'(false_renamed, x7, y6, nil) -> false
   del'(x8, nil) -> false
   equal_sort[a44](eq(x3, y2), true_renamed) -> true | del'(x3, cons(y2, z')) -> true
   equal_sort[a44](eq(x3, y2), true_renamed) -> false | del'(x3, cons(y2, z')) -> del'(x3, z')
   min(x', nil) -> x'
   equal_sort[a44](le(x'', y'), true_renamed) -> true | min(x'', cons(y', z)) -> min(x'', z)
   equal_sort[a44](le(x'', y'), true_renamed) -> false | min(x'', cons(y', z)) -> min(y', z)
   eq(0, 0) -> true_renamed
   eq(0, s(y3)) -> false_renamed
   eq(s(x4), 0) -> false_renamed
   eq(s(x5), s(y4)) -> eq(x5, y4)
   if2(true_renamed, x6, y5, xs'') -> xs''
   if2(false_renamed, x7, y6, cons(y2, z')) -> cons(y6, if2(eq(x7, y2), x7, y2, z'))
   if2(false_renamed, x7, y6, nil) -> cons(y6, nil)
   del(x8, nil) -> nil
   equal_sort[a44](eq(x3, y2), true_renamed) -> true | del(x3, cons(y2, z')) -> z'
   equal_sort[a44](eq(x3, y2), true_renamed) -> false | del(x3, cons(y2, z')) -> cons(y2, del(x3, z'))
   le(0, y7) -> true_renamed
   le(s(x9), 0) -> false_renamed
   le(s(x10), s(y8)) -> le(x10, y8)
   if1(true_renamed, x1, y'', nil) -> x1
   if1(true_renamed, x1, y'', cons(y', z)) -> if1(le(x1, y'), x1, y', z)
   if1(false_renamed, x2, y1, nil) -> y1
   if1(false_renamed, x2, y1, cons(y', z)) -> if1(le(y1, y'), y1, y', z)

using the following formula:
x:sort[a0],y:sort[a33].del'(min(x, y), cons(x, y))=true

could be successfully shown:
(0) Formula
(1) Conditional Evaluation [EQUIVALENT, 0 ms]
(2) AND
    (3) Formula
        (4) Symbolic evaluation [EQUIVALENT, 0 ms]
        (5) YES
    (6) Formula
        (7) Hypothesis Lifting [EQUIVALENT, 0 ms]
        (8) Formula
        (9) Induction by algorithm [EQUIVALENT, 0 ms]
        (10) AND
            (11) Formula
                (12) Symbolic evaluation [EQUIVALENT, 0 ms]
                (13) Formula
                (14) Induction by data structure [EQUIVALENT, 0 ms]
                (15) AND
                    (16) Formula
                        (17) Symbolic evaluation [EQUIVALENT, 0 ms]
                        (18) YES
                    (19) Formula
                        (20) Symbolic evaluation under hypothesis [EQUIVALENT, 0 ms]
                        (21) YES
            (22) Formula
                (23) Conditional Evaluation [EQUIVALENT, 0 ms]
                (24) Formula
                (25) Conditional Evaluation [EQUIVALENT, 0 ms]
                (26) Formula
                (27) Conditional Evaluation [EQUIVALENT, 0 ms]
                (28) AND
                    (29) Formula
                        (30) Symbolic evaluation [EQUIVALENT, 0 ms]
                        (31) YES
                    (32) Formula
                        (33) Symbolic evaluation under hypothesis [EQUIVALENT, 0 ms]
                        (34) YES
            (35) Formula
                (36) Conditional Evaluation [EQUIVALENT, 0 ms]
                (37) Formula
                (38) Conditional Evaluation [EQUIVALENT, 0 ms]
                (39) Formula
                (40) Conditional Evaluation [EQUIVALENT, 0 ms]
                (41) AND
                    (42) Formula
                        (43) Symbolic evaluation [EQUIVALENT, 0 ms]
                        (44) YES
                    (45) Formula
                        (46) Symbolic evaluation under hypothesis [EQUIVALENT, 0 ms]
                        (47) YES


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

(0)
Obligation:
Formula:
x:sort[a0],y:sort[a33].del'(min(x, y), cons(x, y))=true

There are no hypotheses.




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

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

Hypotheses:
x:sort[a0],y:sort[a33].equal_sort[a44](eq(min(x, y), x), true_renamed)=true





Formula:
x:sort[a0],y:sort[a33].del'(min(x, y), y)=true

Hypotheses:
x:sort[a0],y:sort[a33].equal_sort[a44](eq(min(x, y), x), true_renamed)=false






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

(2)
Complex Obligation (AND)

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

(3)
Obligation:
Formula:
true=true

Hypotheses:
x:sort[a0],y:sort[a33].equal_sort[a44](eq(min(x, y), x), true_renamed)=true




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

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

(5)
YES

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

(6)
Obligation:
Formula:
x:sort[a0],y:sort[a33].del'(min(x, y), y)=true

Hypotheses:
x:sort[a0],y:sort[a33].equal_sort[a44](eq(min(x, y), x), true_renamed)=false




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

(7) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
x:sort[a0],y:sort[a33].(equal_sort[a44](eq(min(x, y), x), true_renamed)=false->del'(min(x, y), y)=true)

There are no hypotheses.




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

(8)
Obligation:
Formula:
x:sort[a0],y:sort[a33].(equal_sort[a44](eq(min(x, y), x), true_renamed)=false->del'(min(x, y), y)=true)

There are no hypotheses.




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

(9) Induction by algorithm (EQUIVALENT)
Induction by algorithm min(x, y) generates the following cases:

1. Base Case:
Formula:
x':sort[a0].(equal_sort[a44](eq(min(x', nil), x'), true_renamed)=false->del'(min(x', nil), nil)=true)

There are no hypotheses.





1. Step Case:
Formula:
x'':sort[a0],y':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(x'', cons(y', z)), x''), true_renamed)=false->del'(min(x'', cons(y', z)), cons(y', z))=true)

Hypotheses:
x'':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(x'', z), x''), true_renamed)=false->del'(min(x'', z), z)=true)
x'':sort[a0],y':sort[a0].equal_sort[a44](le(x'', y'), true_renamed)=true





2. Step Case:
Formula:
x'':sort[a0],y':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(x'', cons(y', z)), x''), true_renamed)=false->del'(min(x'', cons(y', z)), cons(y', z))=true)

Hypotheses:
y':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(y', z), y'), true_renamed)=false->del'(min(y', z), z)=true)
x'':sort[a0],y':sort[a0].equal_sort[a44](le(x'', y'), true_renamed)=false






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

(10)
Complex Obligation (AND)

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

(11)
Obligation:
Formula:
x':sort[a0].(equal_sort[a44](eq(min(x', nil), x'), true_renamed)=false->del'(min(x', nil), nil)=true)

There are no hypotheses.




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

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

(13)
Obligation:
Formula:
x':sort[a0].~(equal_sort[a44](eq(x', x'), true_renamed)=false)

There are no hypotheses.




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

(14) Induction by data structure (EQUIVALENT)
Induction by data structure sort[a0] generates the following cases:



1. Base Case:
Formula:
~(equal_sort[a44](eq(0, 0), true_renamed)=false)

There are no hypotheses.





1. Step Case:
Formula:
n:sort[a0].~(equal_sort[a44](eq(s(n), s(n)), true_renamed)=false)

Hypotheses:
n:sort[a0].~(equal_sort[a44](eq(n, n), true_renamed)=false)






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

(15)
Complex Obligation (AND)

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

(16)
Obligation:
Formula:
~(equal_sort[a44](eq(0, 0), true_renamed)=false)

There are no hypotheses.




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

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

(18)
YES

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

(19)
Obligation:
Formula:
n:sort[a0].~(equal_sort[a44](eq(s(n), s(n)), true_renamed)=false)

Hypotheses:
n:sort[a0].~(equal_sort[a44](eq(n, n), true_renamed)=false)




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

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

n:sort[a0].~(equal_sort[a44](eq(n, n), true_renamed)=false)

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

(21)
YES

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

(22)
Obligation:
Formula:
x'':sort[a0],y':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(x'', cons(y', z)), x''), true_renamed)=false->del'(min(x'', cons(y', z)), cons(y', z))=true)

Hypotheses:
x'':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(x'', z), x''), true_renamed)=false->del'(min(x'', z), z)=true)
x'':sort[a0],y':sort[a0].equal_sort[a44](le(x'', y'), true_renamed)=true




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

(23) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
x'':sort[a0],z:sort[a33],y':sort[a0].(equal_sort[a44](eq(min(x'', z), x''), true_renamed)=false->del'(min(x'', cons(y', z)), cons(y', z))=true)

Hypotheses:
x'':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(x'', z), x''), true_renamed)=false->del'(min(x'', z), z)=true)
x'':sort[a0],y':sort[a0].equal_sort[a44](le(x'', y'), true_renamed)=true






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

(24)
Obligation:
Formula:
x'':sort[a0],z:sort[a33],y':sort[a0].(equal_sort[a44](eq(min(x'', z), x''), true_renamed)=false->del'(min(x'', cons(y', z)), cons(y', z))=true)

Hypotheses:
x'':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(x'', z), x''), true_renamed)=false->del'(min(x'', z), z)=true)
x'':sort[a0],y':sort[a0].equal_sort[a44](le(x'', y'), true_renamed)=true




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

(25) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
x'':sort[a0],z:sort[a33],y':sort[a0].(equal_sort[a44](eq(min(x'', z), x''), true_renamed)=false->del'(min(x'', z), cons(y', z))=true)

Hypotheses:
x'':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(x'', z), x''), true_renamed)=false->del'(min(x'', z), z)=true)
x'':sort[a0],y':sort[a0].equal_sort[a44](le(x'', y'), true_renamed)=true






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

(26)
Obligation:
Formula:
x'':sort[a0],z:sort[a33],y':sort[a0].(equal_sort[a44](eq(min(x'', z), x''), true_renamed)=false->del'(min(x'', z), cons(y', z))=true)

Hypotheses:
x'':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(x'', z), x''), true_renamed)=false->del'(min(x'', z), z)=true)
x'':sort[a0],y':sort[a0].equal_sort[a44](le(x'', y'), true_renamed)=true




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

(27) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
x'':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(x'', z), x''), true_renamed)=false->true=true)

Hypotheses:
x'':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(x'', z), x''), true_renamed)=false->del'(min(x'', z), z)=true)
x'':sort[a0],y':sort[a0].equal_sort[a44](le(x'', y'), true_renamed)=true
x'':sort[a0],z:sort[a33],y':sort[a0].equal_sort[a44](eq(min(x'', z), y'), true_renamed)=true





Formula:
x'':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(x'', z), x''), true_renamed)=false->del'(min(x'', z), z)=true)

Hypotheses:
x'':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(x'', z), x''), true_renamed)=false->del'(min(x'', z), z)=true)
x'':sort[a0],y':sort[a0].equal_sort[a44](le(x'', y'), true_renamed)=true
x'':sort[a0],z:sort[a33],y':sort[a0].equal_sort[a44](eq(min(x'', z), y'), true_renamed)=false






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

(28)
Complex Obligation (AND)

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

(29)
Obligation:
Formula:
x'':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(x'', z), x''), true_renamed)=false->true=true)

Hypotheses:
x'':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(x'', z), x''), true_renamed)=false->del'(min(x'', z), z)=true)
x'':sort[a0],y':sort[a0].equal_sort[a44](le(x'', y'), true_renamed)=true
x'':sort[a0],z:sort[a33],y':sort[a0].equal_sort[a44](eq(min(x'', z), y'), true_renamed)=true




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

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

(31)
YES

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

(32)
Obligation:
Formula:
x'':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(x'', z), x''), true_renamed)=false->del'(min(x'', z), z)=true)

Hypotheses:
x'':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(x'', z), x''), true_renamed)=false->del'(min(x'', z), z)=true)
x'':sort[a0],y':sort[a0].equal_sort[a44](le(x'', y'), true_renamed)=true
x'':sort[a0],z:sort[a33],y':sort[a0].equal_sort[a44](eq(min(x'', z), y'), true_renamed)=false




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

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

x'':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(x'', z), x''), true_renamed)=false->del'(min(x'', z), z)=true)

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

(34)
YES

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

(35)
Obligation:
Formula:
x'':sort[a0],y':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(x'', cons(y', z)), x''), true_renamed)=false->del'(min(x'', cons(y', z)), cons(y', z))=true)

Hypotheses:
y':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(y', z), y'), true_renamed)=false->del'(min(y', z), z)=true)
x'':sort[a0],y':sort[a0].equal_sort[a44](le(x'', y'), true_renamed)=false




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

(36) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
y':sort[a0],z:sort[a33],x'':sort[a0].(equal_sort[a44](eq(min(y', z), x''), true_renamed)=false->del'(min(x'', cons(y', z)), cons(y', z))=true)

Hypotheses:
y':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(y', z), y'), true_renamed)=false->del'(min(y', z), z)=true)
x'':sort[a0],y':sort[a0].equal_sort[a44](le(x'', y'), true_renamed)=false






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

(37)
Obligation:
Formula:
y':sort[a0],z:sort[a33],x'':sort[a0].(equal_sort[a44](eq(min(y', z), x''), true_renamed)=false->del'(min(x'', cons(y', z)), cons(y', z))=true)

Hypotheses:
y':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(y', z), y'), true_renamed)=false->del'(min(y', z), z)=true)
x'':sort[a0],y':sort[a0].equal_sort[a44](le(x'', y'), true_renamed)=false




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

(38) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
y':sort[a0],z:sort[a33],x'':sort[a0].(equal_sort[a44](eq(min(y', z), x''), true_renamed)=false->del'(min(y', z), cons(y', z))=true)

Hypotheses:
y':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(y', z), y'), true_renamed)=false->del'(min(y', z), z)=true)
x'':sort[a0],y':sort[a0].equal_sort[a44](le(x'', y'), true_renamed)=false






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

(39)
Obligation:
Formula:
y':sort[a0],z:sort[a33],x'':sort[a0].(equal_sort[a44](eq(min(y', z), x''), true_renamed)=false->del'(min(y', z), cons(y', z))=true)

Hypotheses:
y':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(y', z), y'), true_renamed)=false->del'(min(y', z), z)=true)
x'':sort[a0],y':sort[a0].equal_sort[a44](le(x'', y'), true_renamed)=false




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

(40) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
y':sort[a0],z:sort[a33],x'':sort[a0].(equal_sort[a44](eq(min(y', z), x''), true_renamed)=false->true=true)

Hypotheses:
y':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(y', z), y'), true_renamed)=false->del'(min(y', z), z)=true)
x'':sort[a0],y':sort[a0].equal_sort[a44](le(x'', y'), true_renamed)=false
y':sort[a0],z:sort[a33].equal_sort[a44](eq(min(y', z), y'), true_renamed)=true





Formula:
y':sort[a0],z:sort[a33],x'':sort[a0].(equal_sort[a44](eq(min(y', z), x''), true_renamed)=false->del'(min(y', z), z)=true)

Hypotheses:
y':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(y', z), y'), true_renamed)=false->del'(min(y', z), z)=true)
x'':sort[a0],y':sort[a0].equal_sort[a44](le(x'', y'), true_renamed)=false
y':sort[a0],z:sort[a33].equal_sort[a44](eq(min(y', z), y'), true_renamed)=false






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

(41)
Complex Obligation (AND)

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

(42)
Obligation:
Formula:
y':sort[a0],z:sort[a33],x'':sort[a0].(equal_sort[a44](eq(min(y', z), x''), true_renamed)=false->true=true)

Hypotheses:
y':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(y', z), y'), true_renamed)=false->del'(min(y', z), z)=true)
x'':sort[a0],y':sort[a0].equal_sort[a44](le(x'', y'), true_renamed)=false
y':sort[a0],z:sort[a33].equal_sort[a44](eq(min(y', z), y'), true_renamed)=true




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

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

(44)
YES

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

(45)
Obligation:
Formula:
y':sort[a0],z:sort[a33],x'':sort[a0].(equal_sort[a44](eq(min(y', z), x''), true_renamed)=false->del'(min(y', z), z)=true)

Hypotheses:
y':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(y', z), y'), true_renamed)=false->del'(min(y', z), z)=true)
x'':sort[a0],y':sort[a0].equal_sort[a44](le(x'', y'), true_renamed)=false
y':sort[a0],z:sort[a33].equal_sort[a44](eq(min(y', z), y'), true_renamed)=false




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

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

y':sort[a0],z:sort[a33].(equal_sort[a44](eq(min(y', z), y'), true_renamed)=false->del'(min(y', z), z)=true)
y':sort[a0],z:sort[a33].equal_sort[a44](eq(min(y', z), y'), true_renamed)=false

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

(47)
YES

(41) Complex Obligation (AND)

(42) Obligation:

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

min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)
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))
del(x, nil) → nil
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

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

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

(43) PisEmptyProof (EQUIVALENT transformation)

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

(44) YES

(45) Obligation:

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

del'(x3, cons(y2, z')) → if2'(eq(x3, y2), x3, y2, z')
if2'(true_renamed, x6, y5, xs'') → true
if2'(false_renamed, x7, y6, xs1) → del'(x7, xs1)
del'(x8, nil) → false
min(x', nil) → x'
min(x'', cons(y', z)) → if1(le(x'', y'), x'', y', z)
if1(true_renamed, x1, y'', xs) → min(x1, xs)
if1(false_renamed, x2, y1, xs') → min(y1, xs')
del(x3, cons(y2, z')) → if2(eq(x3, y2), x3, y2, z')
eq(0, 0) → true_renamed
eq(0, s(y3)) → false_renamed
eq(s(x4), 0) → false_renamed
eq(s(x5), s(y4)) → eq(x5, y4)
if2(true_renamed, x6, y5, xs'') → xs''
if2(false_renamed, x7, y6, xs1) → cons(y6, del(x7, xs1))
del(x8, nil) → nil
le(0, y7) → true_renamed
le(s(x9), 0) → false_renamed
le(s(x10), s(y8)) → le(x10, y8)
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[a0](0, 0) → true
equal_sort[a0](0, s(v25)) → false
equal_sort[a0](s(v26), 0) → false
equal_sort[a0](s(v26), s(v27)) → equal_sort[a0](v26, v27)
equal_sort[a33](nil, nil) → true
equal_sort[a33](nil, cons(v28, v29)) → false
equal_sort[a33](cons(v30, v31), nil) → false
equal_sort[a33](cons(v30, v31), cons(v32, v33)) → and(equal_sort[a0](v30, v32), equal_sort[a33](v31, v33))
equal_sort[a44](true_renamed, true_renamed) → true
equal_sort[a44](true_renamed, false_renamed) → false
equal_sort[a44](false_renamed, true_renamed) → false
equal_sort[a44](false_renamed, false_renamed) → true
equal_sort[a62](witness_sort[a62], witness_sort[a62]) → true

Q is empty.

(46) Overlay + Local Confluence (EQUIVALENT transformation)

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

(47) Obligation:

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

del'(x3, cons(y2, z')) → if2'(eq(x3, y2), x3, y2, z')
if2'(true_renamed, x6, y5, xs'') → true
if2'(false_renamed, x7, y6, xs1) → del'(x7, xs1)
del'(x8, nil) → false
min(x', nil) → x'
min(x'', cons(y', z)) → if1(le(x'', y'), x'', y', z)
if1(true_renamed, x1, y'', xs) → min(x1, xs)
if1(false_renamed, x2, y1, xs') → min(y1, xs')
del(x3, cons(y2, z')) → if2(eq(x3, y2), x3, y2, z')
eq(0, 0) → true_renamed
eq(0, s(y3)) → false_renamed
eq(s(x4), 0) → false_renamed
eq(s(x5), s(y4)) → eq(x5, y4)
if2(true_renamed, x6, y5, xs'') → xs''
if2(false_renamed, x7, y6, xs1) → cons(y6, del(x7, xs1))
del(x8, nil) → nil
le(0, y7) → true_renamed
le(s(x9), 0) → false_renamed
le(s(x10), s(y8)) → le(x10, y8)
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[a0](0, 0) → true
equal_sort[a0](0, s(v25)) → false
equal_sort[a0](s(v26), 0) → false
equal_sort[a0](s(v26), s(v27)) → equal_sort[a0](v26, v27)
equal_sort[a33](nil, nil) → true
equal_sort[a33](nil, cons(v28, v29)) → false
equal_sort[a33](cons(v30, v31), nil) → false
equal_sort[a33](cons(v30, v31), cons(v32, v33)) → and(equal_sort[a0](v30, v32), equal_sort[a33](v31, v33))
equal_sort[a44](true_renamed, true_renamed) → true
equal_sort[a44](true_renamed, false_renamed) → false
equal_sort[a44](false_renamed, true_renamed) → false
equal_sort[a44](false_renamed, false_renamed) → true
equal_sort[a62](witness_sort[a62], witness_sort[a62]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
min(x0, cons(x1, x2))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
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)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a33](nil, nil)
equal_sort[a33](nil, cons(x0, x1))
equal_sort[a33](cons(x0, x1), nil)
equal_sort[a33](cons(x0, x1), cons(x2, x3))
equal_sort[a44](true_renamed, true_renamed)
equal_sort[a44](true_renamed, false_renamed)
equal_sort[a44](false_renamed, true_renamed)
equal_sort[a44](false_renamed, false_renamed)
equal_sort[a62](witness_sort[a62], witness_sort[a62])

(48) DependencyPairsProof (EQUIVALENT transformation)

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

(49) Obligation:

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

DEL'(x3, cons(y2, z')) → IF2'(eq(x3, y2), x3, y2, z')
DEL'(x3, cons(y2, z')) → EQ(x3, y2)
IF2'(false_renamed, x7, y6, xs1) → DEL'(x7, xs1)
MIN(x'', cons(y', z)) → IF1(le(x'', y'), x'', y', z)
MIN(x'', cons(y', z)) → LE(x'', y')
IF1(true_renamed, x1, y'', xs) → MIN(x1, xs)
IF1(false_renamed, x2, y1, xs') → MIN(y1, xs')
DEL(x3, cons(y2, z')) → IF2(eq(x3, y2), x3, y2, z')
DEL(x3, cons(y2, z')) → EQ(x3, y2)
EQ(s(x5), s(y4)) → EQ(x5, y4)
IF2(false_renamed, x7, y6, xs1) → DEL(x7, xs1)
LE(s(x10), s(y8)) → LE(x10, y8)
EQUAL_SORT[A0](s(v26), s(v27)) → EQUAL_SORT[A0](v26, v27)
EQUAL_SORT[A33](cons(v30, v31), cons(v32, v33)) → AND(equal_sort[a0](v30, v32), equal_sort[a33](v31, v33))
EQUAL_SORT[A33](cons(v30, v31), cons(v32, v33)) → EQUAL_SORT[A0](v30, v32)
EQUAL_SORT[A33](cons(v30, v31), cons(v32, v33)) → EQUAL_SORT[A33](v31, v33)

The TRS R consists of the following rules:

del'(x3, cons(y2, z')) → if2'(eq(x3, y2), x3, y2, z')
if2'(true_renamed, x6, y5, xs'') → true
if2'(false_renamed, x7, y6, xs1) → del'(x7, xs1)
del'(x8, nil) → false
min(x', nil) → x'
min(x'', cons(y', z)) → if1(le(x'', y'), x'', y', z)
if1(true_renamed, x1, y'', xs) → min(x1, xs)
if1(false_renamed, x2, y1, xs') → min(y1, xs')
del(x3, cons(y2, z')) → if2(eq(x3, y2), x3, y2, z')
eq(0, 0) → true_renamed
eq(0, s(y3)) → false_renamed
eq(s(x4), 0) → false_renamed
eq(s(x5), s(y4)) → eq(x5, y4)
if2(true_renamed, x6, y5, xs'') → xs''
if2(false_renamed, x7, y6, xs1) → cons(y6, del(x7, xs1))
del(x8, nil) → nil
le(0, y7) → true_renamed
le(s(x9), 0) → false_renamed
le(s(x10), s(y8)) → le(x10, y8)
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[a0](0, 0) → true
equal_sort[a0](0, s(v25)) → false
equal_sort[a0](s(v26), 0) → false
equal_sort[a0](s(v26), s(v27)) → equal_sort[a0](v26, v27)
equal_sort[a33](nil, nil) → true
equal_sort[a33](nil, cons(v28, v29)) → false
equal_sort[a33](cons(v30, v31), nil) → false
equal_sort[a33](cons(v30, v31), cons(v32, v33)) → and(equal_sort[a0](v30, v32), equal_sort[a33](v31, v33))
equal_sort[a44](true_renamed, true_renamed) → true
equal_sort[a44](true_renamed, false_renamed) → false
equal_sort[a44](false_renamed, true_renamed) → false
equal_sort[a44](false_renamed, false_renamed) → true
equal_sort[a62](witness_sort[a62], witness_sort[a62]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
min(x0, cons(x1, x2))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
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)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a33](nil, nil)
equal_sort[a33](nil, cons(x0, x1))
equal_sort[a33](cons(x0, x1), nil)
equal_sort[a33](cons(x0, x1), cons(x2, x3))
equal_sort[a44](true_renamed, true_renamed)
equal_sort[a44](true_renamed, false_renamed)
equal_sort[a44](false_renamed, true_renamed)
equal_sort[a44](false_renamed, false_renamed)
equal_sort[a62](witness_sort[a62], witness_sort[a62])

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

(50) DependencyGraphProof (EQUIVALENT transformation)

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

(51) Complex Obligation (AND)

(52) Obligation:

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

EQUAL_SORT[A0](s(v26), s(v27)) → EQUAL_SORT[A0](v26, v27)

The TRS R consists of the following rules:

del'(x3, cons(y2, z')) → if2'(eq(x3, y2), x3, y2, z')
if2'(true_renamed, x6, y5, xs'') → true
if2'(false_renamed, x7, y6, xs1) → del'(x7, xs1)
del'(x8, nil) → false
min(x', nil) → x'
min(x'', cons(y', z)) → if1(le(x'', y'), x'', y', z)
if1(true_renamed, x1, y'', xs) → min(x1, xs)
if1(false_renamed, x2, y1, xs') → min(y1, xs')
del(x3, cons(y2, z')) → if2(eq(x3, y2), x3, y2, z')
eq(0, 0) → true_renamed
eq(0, s(y3)) → false_renamed
eq(s(x4), 0) → false_renamed
eq(s(x5), s(y4)) → eq(x5, y4)
if2(true_renamed, x6, y5, xs'') → xs''
if2(false_renamed, x7, y6, xs1) → cons(y6, del(x7, xs1))
del(x8, nil) → nil
le(0, y7) → true_renamed
le(s(x9), 0) → false_renamed
le(s(x10), s(y8)) → le(x10, y8)
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[a0](0, 0) → true
equal_sort[a0](0, s(v25)) → false
equal_sort[a0](s(v26), 0) → false
equal_sort[a0](s(v26), s(v27)) → equal_sort[a0](v26, v27)
equal_sort[a33](nil, nil) → true
equal_sort[a33](nil, cons(v28, v29)) → false
equal_sort[a33](cons(v30, v31), nil) → false
equal_sort[a33](cons(v30, v31), cons(v32, v33)) → and(equal_sort[a0](v30, v32), equal_sort[a33](v31, v33))
equal_sort[a44](true_renamed, true_renamed) → true
equal_sort[a44](true_renamed, false_renamed) → false
equal_sort[a44](false_renamed, true_renamed) → false
equal_sort[a44](false_renamed, false_renamed) → true
equal_sort[a62](witness_sort[a62], witness_sort[a62]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
min(x0, cons(x1, x2))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
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)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a33](nil, nil)
equal_sort[a33](nil, cons(x0, x1))
equal_sort[a33](cons(x0, x1), nil)
equal_sort[a33](cons(x0, x1), cons(x2, x3))
equal_sort[a44](true_renamed, true_renamed)
equal_sort[a44](true_renamed, false_renamed)
equal_sort[a44](false_renamed, true_renamed)
equal_sort[a44](false_renamed, false_renamed)
equal_sort[a62](witness_sort[a62], witness_sort[a62])

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

(53) 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.

(54) Obligation:

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

EQUAL_SORT[A0](s(v26), s(v27)) → EQUAL_SORT[A0](v26, v27)

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

del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
min(x0, cons(x1, x2))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
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)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a33](nil, nil)
equal_sort[a33](nil, cons(x0, x1))
equal_sort[a33](cons(x0, x1), nil)
equal_sort[a33](cons(x0, x1), cons(x2, x3))
equal_sort[a44](true_renamed, true_renamed)
equal_sort[a44](true_renamed, false_renamed)
equal_sort[a44](false_renamed, true_renamed)
equal_sort[a44](false_renamed, false_renamed)
equal_sort[a62](witness_sort[a62], witness_sort[a62])

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

(55) 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, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
min(x0, cons(x1, x2))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
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)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a33](nil, nil)
equal_sort[a33](nil, cons(x0, x1))
equal_sort[a33](cons(x0, x1), nil)
equal_sort[a33](cons(x0, x1), cons(x2, x3))
equal_sort[a44](true_renamed, true_renamed)
equal_sort[a44](true_renamed, false_renamed)
equal_sort[a44](false_renamed, true_renamed)
equal_sort[a44](false_renamed, false_renamed)
equal_sort[a62](witness_sort[a62], witness_sort[a62])

(56) Obligation:

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

EQUAL_SORT[A0](s(v26), s(v27)) → EQUAL_SORT[A0](v26, v27)

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

(57) 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[A0](s(v26), s(v27)) → EQUAL_SORT[A0](v26, v27)
The graph contains the following edges 1 > 1, 2 > 2

(58) YES

(59) Obligation:

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

EQUAL_SORT[A33](cons(v30, v31), cons(v32, v33)) → EQUAL_SORT[A33](v31, v33)

The TRS R consists of the following rules:

del'(x3, cons(y2, z')) → if2'(eq(x3, y2), x3, y2, z')
if2'(true_renamed, x6, y5, xs'') → true
if2'(false_renamed, x7, y6, xs1) → del'(x7, xs1)
del'(x8, nil) → false
min(x', nil) → x'
min(x'', cons(y', z)) → if1(le(x'', y'), x'', y', z)
if1(true_renamed, x1, y'', xs) → min(x1, xs)
if1(false_renamed, x2, y1, xs') → min(y1, xs')
del(x3, cons(y2, z')) → if2(eq(x3, y2), x3, y2, z')
eq(0, 0) → true_renamed
eq(0, s(y3)) → false_renamed
eq(s(x4), 0) → false_renamed
eq(s(x5), s(y4)) → eq(x5, y4)
if2(true_renamed, x6, y5, xs'') → xs''
if2(false_renamed, x7, y6, xs1) → cons(y6, del(x7, xs1))
del(x8, nil) → nil
le(0, y7) → true_renamed
le(s(x9), 0) → false_renamed
le(s(x10), s(y8)) → le(x10, y8)
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[a0](0, 0) → true
equal_sort[a0](0, s(v25)) → false
equal_sort[a0](s(v26), 0) → false
equal_sort[a0](s(v26), s(v27)) → equal_sort[a0](v26, v27)
equal_sort[a33](nil, nil) → true
equal_sort[a33](nil, cons(v28, v29)) → false
equal_sort[a33](cons(v30, v31), nil) → false
equal_sort[a33](cons(v30, v31), cons(v32, v33)) → and(equal_sort[a0](v30, v32), equal_sort[a33](v31, v33))
equal_sort[a44](true_renamed, true_renamed) → true
equal_sort[a44](true_renamed, false_renamed) → false
equal_sort[a44](false_renamed, true_renamed) → false
equal_sort[a44](false_renamed, false_renamed) → true
equal_sort[a62](witness_sort[a62], witness_sort[a62]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
min(x0, cons(x1, x2))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
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)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a33](nil, nil)
equal_sort[a33](nil, cons(x0, x1))
equal_sort[a33](cons(x0, x1), nil)
equal_sort[a33](cons(x0, x1), cons(x2, x3))
equal_sort[a44](true_renamed, true_renamed)
equal_sort[a44](true_renamed, false_renamed)
equal_sort[a44](false_renamed, true_renamed)
equal_sort[a44](false_renamed, false_renamed)
equal_sort[a62](witness_sort[a62], witness_sort[a62])

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

(60) 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.

(61) Obligation:

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

EQUAL_SORT[A33](cons(v30, v31), cons(v32, v33)) → EQUAL_SORT[A33](v31, v33)

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

del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
min(x0, cons(x1, x2))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
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)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a33](nil, nil)
equal_sort[a33](nil, cons(x0, x1))
equal_sort[a33](cons(x0, x1), nil)
equal_sort[a33](cons(x0, x1), cons(x2, x3))
equal_sort[a44](true_renamed, true_renamed)
equal_sort[a44](true_renamed, false_renamed)
equal_sort[a44](false_renamed, true_renamed)
equal_sort[a44](false_renamed, false_renamed)
equal_sort[a62](witness_sort[a62], witness_sort[a62])

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

(62) 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, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
min(x0, cons(x1, x2))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
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)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a33](nil, nil)
equal_sort[a33](nil, cons(x0, x1))
equal_sort[a33](cons(x0, x1), nil)
equal_sort[a33](cons(x0, x1), cons(x2, x3))
equal_sort[a44](true_renamed, true_renamed)
equal_sort[a44](true_renamed, false_renamed)
equal_sort[a44](false_renamed, true_renamed)
equal_sort[a44](false_renamed, false_renamed)
equal_sort[a62](witness_sort[a62], witness_sort[a62])

(63) Obligation:

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

EQUAL_SORT[A33](cons(v30, v31), cons(v32, v33)) → EQUAL_SORT[A33](v31, v33)

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

(64) 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](cons(v30, v31), cons(v32, v33)) → EQUAL_SORT[A33](v31, v33)
The graph contains the following edges 1 > 1, 2 > 2

(65) YES

(66) Obligation:

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

LE(s(x10), s(y8)) → LE(x10, y8)

The TRS R consists of the following rules:

del'(x3, cons(y2, z')) → if2'(eq(x3, y2), x3, y2, z')
if2'(true_renamed, x6, y5, xs'') → true
if2'(false_renamed, x7, y6, xs1) → del'(x7, xs1)
del'(x8, nil) → false
min(x', nil) → x'
min(x'', cons(y', z)) → if1(le(x'', y'), x'', y', z)
if1(true_renamed, x1, y'', xs) → min(x1, xs)
if1(false_renamed, x2, y1, xs') → min(y1, xs')
del(x3, cons(y2, z')) → if2(eq(x3, y2), x3, y2, z')
eq(0, 0) → true_renamed
eq(0, s(y3)) → false_renamed
eq(s(x4), 0) → false_renamed
eq(s(x5), s(y4)) → eq(x5, y4)
if2(true_renamed, x6, y5, xs'') → xs''
if2(false_renamed, x7, y6, xs1) → cons(y6, del(x7, xs1))
del(x8, nil) → nil
le(0, y7) → true_renamed
le(s(x9), 0) → false_renamed
le(s(x10), s(y8)) → le(x10, y8)
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[a0](0, 0) → true
equal_sort[a0](0, s(v25)) → false
equal_sort[a0](s(v26), 0) → false
equal_sort[a0](s(v26), s(v27)) → equal_sort[a0](v26, v27)
equal_sort[a33](nil, nil) → true
equal_sort[a33](nil, cons(v28, v29)) → false
equal_sort[a33](cons(v30, v31), nil) → false
equal_sort[a33](cons(v30, v31), cons(v32, v33)) → and(equal_sort[a0](v30, v32), equal_sort[a33](v31, v33))
equal_sort[a44](true_renamed, true_renamed) → true
equal_sort[a44](true_renamed, false_renamed) → false
equal_sort[a44](false_renamed, true_renamed) → false
equal_sort[a44](false_renamed, false_renamed) → true
equal_sort[a62](witness_sort[a62], witness_sort[a62]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
min(x0, cons(x1, x2))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
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)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a33](nil, nil)
equal_sort[a33](nil, cons(x0, x1))
equal_sort[a33](cons(x0, x1), nil)
equal_sort[a33](cons(x0, x1), cons(x2, x3))
equal_sort[a44](true_renamed, true_renamed)
equal_sort[a44](true_renamed, false_renamed)
equal_sort[a44](false_renamed, true_renamed)
equal_sort[a44](false_renamed, false_renamed)
equal_sort[a62](witness_sort[a62], witness_sort[a62])

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

(67) 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.

(68) Obligation:

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

LE(s(x10), s(y8)) → LE(x10, y8)

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

del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
min(x0, cons(x1, x2))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
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)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a33](nil, nil)
equal_sort[a33](nil, cons(x0, x1))
equal_sort[a33](cons(x0, x1), nil)
equal_sort[a33](cons(x0, x1), cons(x2, x3))
equal_sort[a44](true_renamed, true_renamed)
equal_sort[a44](true_renamed, false_renamed)
equal_sort[a44](false_renamed, true_renamed)
equal_sort[a44](false_renamed, false_renamed)
equal_sort[a62](witness_sort[a62], witness_sort[a62])

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

(69) 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, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
min(x0, cons(x1, x2))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
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)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a33](nil, nil)
equal_sort[a33](nil, cons(x0, x1))
equal_sort[a33](cons(x0, x1), nil)
equal_sort[a33](cons(x0, x1), cons(x2, x3))
equal_sort[a44](true_renamed, true_renamed)
equal_sort[a44](true_renamed, false_renamed)
equal_sort[a44](false_renamed, true_renamed)
equal_sort[a44](false_renamed, false_renamed)
equal_sort[a62](witness_sort[a62], witness_sort[a62])

(70) Obligation:

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

LE(s(x10), s(y8)) → LE(x10, y8)

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

(71) 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:

LE(s(x10), s(y8)) → LE(x10, y8)
The graph contains the following edges 1 > 1, 2 > 2

(72) YES

(73) Obligation:

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

EQ(s(x5), s(y4)) → EQ(x5, y4)

The TRS R consists of the following rules:

del'(x3, cons(y2, z')) → if2'(eq(x3, y2), x3, y2, z')
if2'(true_renamed, x6, y5, xs'') → true
if2'(false_renamed, x7, y6, xs1) → del'(x7, xs1)
del'(x8, nil) → false
min(x', nil) → x'
min(x'', cons(y', z)) → if1(le(x'', y'), x'', y', z)
if1(true_renamed, x1, y'', xs) → min(x1, xs)
if1(false_renamed, x2, y1, xs') → min(y1, xs')
del(x3, cons(y2, z')) → if2(eq(x3, y2), x3, y2, z')
eq(0, 0) → true_renamed
eq(0, s(y3)) → false_renamed
eq(s(x4), 0) → false_renamed
eq(s(x5), s(y4)) → eq(x5, y4)
if2(true_renamed, x6, y5, xs'') → xs''
if2(false_renamed, x7, y6, xs1) → cons(y6, del(x7, xs1))
del(x8, nil) → nil
le(0, y7) → true_renamed
le(s(x9), 0) → false_renamed
le(s(x10), s(y8)) → le(x10, y8)
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[a0](0, 0) → true
equal_sort[a0](0, s(v25)) → false
equal_sort[a0](s(v26), 0) → false
equal_sort[a0](s(v26), s(v27)) → equal_sort[a0](v26, v27)
equal_sort[a33](nil, nil) → true
equal_sort[a33](nil, cons(v28, v29)) → false
equal_sort[a33](cons(v30, v31), nil) → false
equal_sort[a33](cons(v30, v31), cons(v32, v33)) → and(equal_sort[a0](v30, v32), equal_sort[a33](v31, v33))
equal_sort[a44](true_renamed, true_renamed) → true
equal_sort[a44](true_renamed, false_renamed) → false
equal_sort[a44](false_renamed, true_renamed) → false
equal_sort[a44](false_renamed, false_renamed) → true
equal_sort[a62](witness_sort[a62], witness_sort[a62]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
min(x0, cons(x1, x2))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
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)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a33](nil, nil)
equal_sort[a33](nil, cons(x0, x1))
equal_sort[a33](cons(x0, x1), nil)
equal_sort[a33](cons(x0, x1), cons(x2, x3))
equal_sort[a44](true_renamed, true_renamed)
equal_sort[a44](true_renamed, false_renamed)
equal_sort[a44](false_renamed, true_renamed)
equal_sort[a44](false_renamed, false_renamed)
equal_sort[a62](witness_sort[a62], witness_sort[a62])

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

(74) 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.

(75) Obligation:

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

EQ(s(x5), s(y4)) → EQ(x5, y4)

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

del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
min(x0, cons(x1, x2))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
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)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a33](nil, nil)
equal_sort[a33](nil, cons(x0, x1))
equal_sort[a33](cons(x0, x1), nil)
equal_sort[a33](cons(x0, x1), cons(x2, x3))
equal_sort[a44](true_renamed, true_renamed)
equal_sort[a44](true_renamed, false_renamed)
equal_sort[a44](false_renamed, true_renamed)
equal_sort[a44](false_renamed, false_renamed)
equal_sort[a62](witness_sort[a62], witness_sort[a62])

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

(76) 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, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
min(x0, cons(x1, x2))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
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)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a33](nil, nil)
equal_sort[a33](nil, cons(x0, x1))
equal_sort[a33](cons(x0, x1), nil)
equal_sort[a33](cons(x0, x1), cons(x2, x3))
equal_sort[a44](true_renamed, true_renamed)
equal_sort[a44](true_renamed, false_renamed)
equal_sort[a44](false_renamed, true_renamed)
equal_sort[a44](false_renamed, false_renamed)
equal_sort[a62](witness_sort[a62], witness_sort[a62])

(77) Obligation:

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

EQ(s(x5), s(y4)) → EQ(x5, y4)

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

(78) 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(x5), s(y4)) → EQ(x5, y4)
The graph contains the following edges 1 > 1, 2 > 2

(79) YES

(80) Obligation:

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

IF2(false_renamed, x7, y6, xs1) → DEL(x7, xs1)
DEL(x3, cons(y2, z')) → IF2(eq(x3, y2), x3, y2, z')

The TRS R consists of the following rules:

del'(x3, cons(y2, z')) → if2'(eq(x3, y2), x3, y2, z')
if2'(true_renamed, x6, y5, xs'') → true
if2'(false_renamed, x7, y6, xs1) → del'(x7, xs1)
del'(x8, nil) → false
min(x', nil) → x'
min(x'', cons(y', z)) → if1(le(x'', y'), x'', y', z)
if1(true_renamed, x1, y'', xs) → min(x1, xs)
if1(false_renamed, x2, y1, xs') → min(y1, xs')
del(x3, cons(y2, z')) → if2(eq(x3, y2), x3, y2, z')
eq(0, 0) → true_renamed
eq(0, s(y3)) → false_renamed
eq(s(x4), 0) → false_renamed
eq(s(x5), s(y4)) → eq(x5, y4)
if2(true_renamed, x6, y5, xs'') → xs''
if2(false_renamed, x7, y6, xs1) → cons(y6, del(x7, xs1))
del(x8, nil) → nil
le(0, y7) → true_renamed
le(s(x9), 0) → false_renamed
le(s(x10), s(y8)) → le(x10, y8)
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[a0](0, 0) → true
equal_sort[a0](0, s(v25)) → false
equal_sort[a0](s(v26), 0) → false
equal_sort[a0](s(v26), s(v27)) → equal_sort[a0](v26, v27)
equal_sort[a33](nil, nil) → true
equal_sort[a33](nil, cons(v28, v29)) → false
equal_sort[a33](cons(v30, v31), nil) → false
equal_sort[a33](cons(v30, v31), cons(v32, v33)) → and(equal_sort[a0](v30, v32), equal_sort[a33](v31, v33))
equal_sort[a44](true_renamed, true_renamed) → true
equal_sort[a44](true_renamed, false_renamed) → false
equal_sort[a44](false_renamed, true_renamed) → false
equal_sort[a44](false_renamed, false_renamed) → true
equal_sort[a62](witness_sort[a62], witness_sort[a62]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
min(x0, cons(x1, x2))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
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)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a33](nil, nil)
equal_sort[a33](nil, cons(x0, x1))
equal_sort[a33](cons(x0, x1), nil)
equal_sort[a33](cons(x0, x1), cons(x2, x3))
equal_sort[a44](true_renamed, true_renamed)
equal_sort[a44](true_renamed, false_renamed)
equal_sort[a44](false_renamed, true_renamed)
equal_sort[a44](false_renamed, false_renamed)
equal_sort[a62](witness_sort[a62], witness_sort[a62])

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

(81) 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.

(82) Obligation:

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

IF2(false_renamed, x7, y6, xs1) → DEL(x7, xs1)
DEL(x3, cons(y2, z')) → IF2(eq(x3, y2), x3, y2, z')

The TRS R consists of the following rules:

eq(0, 0) → true_renamed
eq(0, s(y3)) → false_renamed
eq(s(x4), 0) → false_renamed
eq(s(x5), s(y4)) → eq(x5, y4)

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
min(x0, cons(x1, x2))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
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)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a33](nil, nil)
equal_sort[a33](nil, cons(x0, x1))
equal_sort[a33](cons(x0, x1), nil)
equal_sort[a33](cons(x0, x1), cons(x2, x3))
equal_sort[a44](true_renamed, true_renamed)
equal_sort[a44](true_renamed, false_renamed)
equal_sort[a44](false_renamed, true_renamed)
equal_sort[a44](false_renamed, false_renamed)
equal_sort[a62](witness_sort[a62], witness_sort[a62])

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

(83) 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, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
min(x0, cons(x1, x2))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
del(x0, cons(x1, x2))
if2(true_renamed, x0, x1, x2)
if2(false_renamed, x0, x1, x2)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a33](nil, nil)
equal_sort[a33](nil, cons(x0, x1))
equal_sort[a33](cons(x0, x1), nil)
equal_sort[a33](cons(x0, x1), cons(x2, x3))
equal_sort[a44](true_renamed, true_renamed)
equal_sort[a44](true_renamed, false_renamed)
equal_sort[a44](false_renamed, true_renamed)
equal_sort[a44](false_renamed, false_renamed)
equal_sort[a62](witness_sort[a62], witness_sort[a62])

(84) Obligation:

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

IF2(false_renamed, x7, y6, xs1) → DEL(x7, xs1)
DEL(x3, cons(y2, z')) → IF2(eq(x3, y2), x3, y2, z')

The TRS R consists of the following rules:

eq(0, 0) → true_renamed
eq(0, s(y3)) → false_renamed
eq(s(x4), 0) → false_renamed
eq(s(x5), s(y4)) → eq(x5, y4)

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.

(85) 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(x3, cons(y2, z')) → IF2(eq(x3, y2), x3, y2, z')
The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4
IF2(false_renamed, x7, y6, xs1) → DEL(x7, xs1)
The graph contains the following edges 2 >= 1, 4 >= 2

(86) YES

(87) Obligation:

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

IF1(true_renamed, x1, y'', xs) → MIN(x1, xs)
MIN(x'', cons(y', z)) → IF1(le(x'', y'), x'', y', z)
IF1(false_renamed, x2, y1, xs') → MIN(y1, xs')

The TRS R consists of the following rules:

del'(x3, cons(y2, z')) → if2'(eq(x3, y2), x3, y2, z')
if2'(true_renamed, x6, y5, xs'') → true
if2'(false_renamed, x7, y6, xs1) → del'(x7, xs1)
del'(x8, nil) → false
min(x', nil) → x'
min(x'', cons(y', z)) → if1(le(x'', y'), x'', y', z)
if1(true_renamed, x1, y'', xs) → min(x1, xs)
if1(false_renamed, x2, y1, xs') → min(y1, xs')
del(x3, cons(y2, z')) → if2(eq(x3, y2), x3, y2, z')
eq(0, 0) → true_renamed
eq(0, s(y3)) → false_renamed
eq(s(x4), 0) → false_renamed
eq(s(x5), s(y4)) → eq(x5, y4)
if2(true_renamed, x6, y5, xs'') → xs''
if2(false_renamed, x7, y6, xs1) → cons(y6, del(x7, xs1))
del(x8, nil) → nil
le(0, y7) → true_renamed
le(s(x9), 0) → false_renamed
le(s(x10), s(y8)) → le(x10, y8)
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[a0](0, 0) → true
equal_sort[a0](0, s(v25)) → false
equal_sort[a0](s(v26), 0) → false
equal_sort[a0](s(v26), s(v27)) → equal_sort[a0](v26, v27)
equal_sort[a33](nil, nil) → true
equal_sort[a33](nil, cons(v28, v29)) → false
equal_sort[a33](cons(v30, v31), nil) → false
equal_sort[a33](cons(v30, v31), cons(v32, v33)) → and(equal_sort[a0](v30, v32), equal_sort[a33](v31, v33))
equal_sort[a44](true_renamed, true_renamed) → true
equal_sort[a44](true_renamed, false_renamed) → false
equal_sort[a44](false_renamed, true_renamed) → false
equal_sort[a44](false_renamed, false_renamed) → true
equal_sort[a62](witness_sort[a62], witness_sort[a62]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
min(x0, cons(x1, x2))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
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)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a33](nil, nil)
equal_sort[a33](nil, cons(x0, x1))
equal_sort[a33](cons(x0, x1), nil)
equal_sort[a33](cons(x0, x1), cons(x2, x3))
equal_sort[a44](true_renamed, true_renamed)
equal_sort[a44](true_renamed, false_renamed)
equal_sort[a44](false_renamed, true_renamed)
equal_sort[a44](false_renamed, false_renamed)
equal_sort[a62](witness_sort[a62], witness_sort[a62])

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

(88) 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.

(89) Obligation:

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

IF1(true_renamed, x1, y'', xs) → MIN(x1, xs)
MIN(x'', cons(y', z)) → IF1(le(x'', y'), x'', y', z)
IF1(false_renamed, x2, y1, xs') → MIN(y1, xs')

The TRS R consists of the following rules:

le(0, y7) → true_renamed
le(s(x9), 0) → false_renamed
le(s(x10), s(y8)) → le(x10, y8)

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
min(x0, cons(x1, x2))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
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)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a33](nil, nil)
equal_sort[a33](nil, cons(x0, x1))
equal_sort[a33](cons(x0, x1), nil)
equal_sort[a33](cons(x0, x1), cons(x2, x3))
equal_sort[a44](true_renamed, true_renamed)
equal_sort[a44](true_renamed, false_renamed)
equal_sort[a44](false_renamed, true_renamed)
equal_sort[a44](false_renamed, false_renamed)
equal_sort[a62](witness_sort[a62], witness_sort[a62])

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

(90) 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, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
min(x0, cons(x1, x2))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
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)
del(x0, nil)
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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a33](nil, nil)
equal_sort[a33](nil, cons(x0, x1))
equal_sort[a33](cons(x0, x1), nil)
equal_sort[a33](cons(x0, x1), cons(x2, x3))
equal_sort[a44](true_renamed, true_renamed)
equal_sort[a44](true_renamed, false_renamed)
equal_sort[a44](false_renamed, true_renamed)
equal_sort[a44](false_renamed, false_renamed)
equal_sort[a62](witness_sort[a62], witness_sort[a62])

(91) Obligation:

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

IF1(true_renamed, x1, y'', xs) → MIN(x1, xs)
MIN(x'', cons(y', z)) → IF1(le(x'', y'), x'', y', z)
IF1(false_renamed, x2, y1, xs') → MIN(y1, xs')

The TRS R consists of the following rules:

le(0, y7) → true_renamed
le(s(x9), 0) → false_renamed
le(s(x10), s(y8)) → le(x10, y8)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

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

(92) 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:

MIN(x'', cons(y', z)) → IF1(le(x'', y'), x'', y', z)
The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4
IF1(true_renamed, x1, y'', xs) → MIN(x1, xs)
The graph contains the following edges 2 >= 1, 4 >= 2
IF1(false_renamed, x2, y1, xs') → MIN(y1, xs')
The graph contains the following edges 3 >= 1, 4 >= 2

(93) YES

(94) Obligation:

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

IF2'(false_renamed, x7, y6, xs1) → DEL'(x7, xs1)
DEL'(x3, cons(y2, z')) → IF2'(eq(x3, y2), x3, y2, z')

The TRS R consists of the following rules:

del'(x3, cons(y2, z')) → if2'(eq(x3, y2), x3, y2, z')
if2'(true_renamed, x6, y5, xs'') → true
if2'(false_renamed, x7, y6, xs1) → del'(x7, xs1)
del'(x8, nil) → false
min(x', nil) → x'
min(x'', cons(y', z)) → if1(le(x'', y'), x'', y', z)
if1(true_renamed, x1, y'', xs) → min(x1, xs)
if1(false_renamed, x2, y1, xs') → min(y1, xs')
del(x3, cons(y2, z')) → if2(eq(x3, y2), x3, y2, z')
eq(0, 0) → true_renamed
eq(0, s(y3)) → false_renamed
eq(s(x4), 0) → false_renamed
eq(s(x5), s(y4)) → eq(x5, y4)
if2(true_renamed, x6, y5, xs'') → xs''
if2(false_renamed, x7, y6, xs1) → cons(y6, del(x7, xs1))
del(x8, nil) → nil
le(0, y7) → true_renamed
le(s(x9), 0) → false_renamed
le(s(x10), s(y8)) → le(x10, y8)
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[a0](0, 0) → true
equal_sort[a0](0, s(v25)) → false
equal_sort[a0](s(v26), 0) → false
equal_sort[a0](s(v26), s(v27)) → equal_sort[a0](v26, v27)
equal_sort[a33](nil, nil) → true
equal_sort[a33](nil, cons(v28, v29)) → false
equal_sort[a33](cons(v30, v31), nil) → false
equal_sort[a33](cons(v30, v31), cons(v32, v33)) → and(equal_sort[a0](v30, v32), equal_sort[a33](v31, v33))
equal_sort[a44](true_renamed, true_renamed) → true
equal_sort[a44](true_renamed, false_renamed) → false
equal_sort[a44](false_renamed, true_renamed) → false
equal_sort[a44](false_renamed, false_renamed) → true
equal_sort[a62](witness_sort[a62], witness_sort[a62]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
min(x0, cons(x1, x2))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
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)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a33](nil, nil)
equal_sort[a33](nil, cons(x0, x1))
equal_sort[a33](cons(x0, x1), nil)
equal_sort[a33](cons(x0, x1), cons(x2, x3))
equal_sort[a44](true_renamed, true_renamed)
equal_sort[a44](true_renamed, false_renamed)
equal_sort[a44](false_renamed, true_renamed)
equal_sort[a44](false_renamed, false_renamed)
equal_sort[a62](witness_sort[a62], witness_sort[a62])

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

(95) 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.

(96) Obligation:

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

IF2'(false_renamed, x7, y6, xs1) → DEL'(x7, xs1)
DEL'(x3, cons(y2, z')) → IF2'(eq(x3, y2), x3, y2, z')

The TRS R consists of the following rules:

eq(0, 0) → true_renamed
eq(0, s(y3)) → false_renamed
eq(s(x4), 0) → false_renamed
eq(s(x5), s(y4)) → eq(x5, y4)

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
min(x0, cons(x1, x2))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
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)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a33](nil, nil)
equal_sort[a33](nil, cons(x0, x1))
equal_sort[a33](cons(x0, x1), nil)
equal_sort[a33](cons(x0, x1), cons(x2, x3))
equal_sort[a44](true_renamed, true_renamed)
equal_sort[a44](true_renamed, false_renamed)
equal_sort[a44](false_renamed, true_renamed)
equal_sort[a44](false_renamed, false_renamed)
equal_sort[a62](witness_sort[a62], witness_sort[a62])

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

(97) 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, cons(x1, x2))
if2'(true_renamed, x0, x1, x2)
if2'(false_renamed, x0, x1, x2)
del'(x0, nil)
min(x0, nil)
min(x0, cons(x1, x2))
if1(true_renamed, x0, x1, x2)
if1(false_renamed, x0, x1, x2)
del(x0, cons(x1, x2))
if2(true_renamed, x0, x1, x2)
if2(false_renamed, x0, x1, x2)
del(x0, nil)
le(0, x0)
le(s(x0), 0)
le(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a33](nil, nil)
equal_sort[a33](nil, cons(x0, x1))
equal_sort[a33](cons(x0, x1), nil)
equal_sort[a33](cons(x0, x1), cons(x2, x3))
equal_sort[a44](true_renamed, true_renamed)
equal_sort[a44](true_renamed, false_renamed)
equal_sort[a44](false_renamed, true_renamed)
equal_sort[a44](false_renamed, false_renamed)
equal_sort[a62](witness_sort[a62], witness_sort[a62])

(98) Obligation:

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

IF2'(false_renamed, x7, y6, xs1) → DEL'(x7, xs1)
DEL'(x3, cons(y2, z')) → IF2'(eq(x3, y2), x3, y2, z')

The TRS R consists of the following rules:

eq(0, 0) → true_renamed
eq(0, s(y3)) → false_renamed
eq(s(x4), 0) → false_renamed
eq(s(x5), s(y4)) → eq(x5, y4)

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.

(99) 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'(x3, cons(y2, z')) → IF2'(eq(x3, y2), x3, y2, z')
The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4
IF2'(false_renamed, x7, y6, xs1) → DEL'(x7, xs1)
The graph contains the following edges 2 >= 1, 4 >= 2