Termination w.r.t. Q proof of AProVE_09_Inductive

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

0 QTRS

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

↳2 QTRS

↳3 DependencyPairsProof (⇔, 0 ms)

↳4 QDP

↳5 DependencyGraphProof (⇔, 0 ms)

↳6 AND

    ↳7 QDP

        ↳8 UsableRulesProof (⇔, 0 ms)

        ↳9 QDP

        ↳10 QReductionProof (⇔, 0 ms)

        ↳11 QDP

        ↳12 QDPSizeChangeProof (⇔, 0 ms)

        ↳13 YES

    ↳14 QDP

        ↳15 UsableRulesProof (⇔, 0 ms)

        ↳16 QDP

        ↳17 QReductionProof (⇔, 0 ms)

        ↳18 QDP

        ↳19 QDPSizeChangeProof (⇔, 0 ms)

        ↳20 YES

    ↳21 QDP

        ↳22 UsableRulesProof (⇔, 0 ms)

        ↳23 QDP

        ↳24 QReductionProof (⇔, 0 ms)

        ↳25 QDP

        ↳26 QDPSizeChangeProof (⇔, 0 ms)

        ↳27 YES

    ↳28 QDP

        ↳29 UsableRulesProof (⇔, 0 ms)

        ↳30 QDP

        ↳31 QReductionProof (⇔, 0 ms)

        ↳32 QDP

        ↳33 TransformationProof (⇔, 0 ms)

        ↳34 QDP

        ↳35 TransformationProof (⇔, 0 ms)

        ↳36 QDP

        ↳37 DependencyGraphProof (⇔, 0 ms)

        ↳38 QDP

        ↳39 TransformationProof (⇔, 0 ms)

        ↳40 QDP

        ↳41 DependencyGraphProof (⇔, 0 ms)

        ↳42 QDP

        ↳43 TransformationProof (⇔, 0 ms)

        ↳44 QDP

        ↳45 TransformationProof (⇔, 0 ms)

        ↳46 QDP

        ↳47 TransformationProof (⇔, 0 ms)

        ↳48 QDP

        ↳49 TransformationProof (⇔, 0 ms)

        ↳50 QDP

        ↳51 TransformationProof (⇔, 0 ms)

        ↳52 QDP

        ↳53 TransformationProof (⇔, 0 ms)

        ↳54 QDP

        ↳55 TransformationProof (⇔, 0 ms)

        ↳56 QDP

        ↳57 TransformationProof (⇔, 0 ms)

        ↳58 QDP

        ↳59 Induction-Processor (⇒, 14 ms)

        ↳60 AND

            ↳61 QDP

                ↳62 DependencyGraphProof (⇔, 0 ms)

                ↳63 QDP

                ↳64 Induction-Processor (⇒, 7 ms)

                ↳65 AND

                    ↳66 QDP

                        ↳67 DependencyGraphProof (⇔, 0 ms)

                        ↳68 TRUE

                    ↳69 QTRS

                        ↳70 QTRSRRRProof (⇔, 0 ms)

                        ↳71 QTRS

                        ↳72 QTRSRRRProof (⇔, 0 ms)

                        ↳73 QTRS

                        ↳74 RisEmptyProof (⇔, 0 ms)

                        ↳75 YES

            ↳76 QTRS

                ↳77 QTRSRRRProof (⇔, 4 ms)

                ↳78 QTRS

                ↳79 QTRSRRRProof (⇔, 0 ms)

                ↳80 QTRS

                ↳81 RisEmptyProof (⇔, 0 ms)

                ↳82 YES

(0) Obligation:

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

minus(s(x), y) → if(gt(s(x), y), x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gcd(x, y) → if1(ge(x, y), x, y)
if1(true, x, y) → if2(gt(y, 0), x, y)
if1(false, x, y) → if3(gt(x, 0), x, y)
if2(true, x, y) → gcd(minus(x, y), y)
if2(false, x, y) → x
if3(true, x, y) → gcd(x, minus(y, x))
if3(false, x, y) → y
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

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

(2) Obligation:

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

minus(s(x), y) → if(gt(s(x), y), x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gcd(x, y) → if1(ge(x, y), x, y)
if1(true, x, y) → if2(gt(y, 0), x, y)
if1(false, x, y) → if3(gt(x, 0), x, y)
if2(true, x, y) → gcd(minus(x, y), y)
if2(false, x, y) → x
if3(true, x, y) → gcd(x, minus(y, x))
if3(false, x, y) → y
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gcd(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

MINUS(s(x), y) → IF(gt(s(x), y), x, y)
MINUS(s(x), y) → GT(s(x), y)
IF(true, x, y) → MINUS(x, y)
GCD(x, y) → IF1(ge(x, y), x, y)
GCD(x, y) → GE(x, y)
IF1(true, x, y) → IF2(gt(y, 0), x, y)
IF1(true, x, y) → GT(y, 0)
IF1(false, x, y) → IF3(gt(x, 0), x, y)
IF1(false, x, y) → GT(x, 0)
IF2(true, x, y) → GCD(minus(x, y), y)
IF2(true, x, y) → MINUS(x, y)
IF3(true, x, y) → GCD(x, minus(y, x))
IF3(true, x, y) → MINUS(y, x)
GT(s(x), s(y)) → GT(x, y)
GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

minus(s(x), y) → if(gt(s(x), y), x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gcd(x, y) → if1(ge(x, y), x, y)
if1(true, x, y) → if2(gt(y, 0), x, y)
if1(false, x, y) → if3(gt(x, 0), x, y)
if2(true, x, y) → gcd(minus(x, y), y)
if2(false, x, y) → x
if3(true, x, y) → gcd(x, minus(y, x))
if3(false, x, y) → y
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gcd(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 6 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

minus(s(x), y) → if(gt(s(x), y), x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gcd(x, y) → if1(ge(x, y), x, y)
if1(true, x, y) → if2(gt(y, 0), x, y)
if1(false, x, y) → if3(gt(x, 0), x, y)
if2(true, x, y) → gcd(minus(x, y), y)
if2(false, x, y) → x
if3(true, x, y) → gcd(x, minus(y, x))
if3(false, x, y) → y
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gcd(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(8) UsableRulesProof (EQUIVALENT transformation)

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

(9) Obligation:

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

GE(s(x), s(y)) → GE(x, y)

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

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gcd(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(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].

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gcd(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

(11) Obligation:

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

GE(s(x), s(y)) → GE(x, y)

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

(12) QDPSizeChangeProof (EQUIVALENT transformation)

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

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

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

(13) YES

(14) Obligation:

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

GT(s(x), s(y)) → GT(x, y)

The TRS R consists of the following rules:

minus(s(x), y) → if(gt(s(x), y), x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gcd(x, y) → if1(ge(x, y), x, y)
if1(true, x, y) → if2(gt(y, 0), x, y)
if1(false, x, y) → if3(gt(x, 0), x, y)
if2(true, x, y) → gcd(minus(x, y), y)
if2(false, x, y) → x
if3(true, x, y) → gcd(x, minus(y, x))
if3(false, x, y) → y
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gcd(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(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:

GT(s(x), s(y)) → GT(x, y)

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

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gcd(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(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].

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gcd(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

(18) Obligation:

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

GT(s(x), s(y)) → GT(x, y)

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

(19) QDPSizeChangeProof (EQUIVALENT transformation)

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

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

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

(20) YES

(21) Obligation:

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

IF(true, x, y) → MINUS(x, y)
MINUS(s(x), y) → IF(gt(s(x), y), x, y)

The TRS R consists of the following rules:

minus(s(x), y) → if(gt(s(x), y), x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gcd(x, y) → if1(ge(x, y), x, y)
if1(true, x, y) → if2(gt(y, 0), x, y)
if1(false, x, y) → if3(gt(x, 0), x, y)
if2(true, x, y) → gcd(minus(x, y), y)
if2(false, x, y) → x
if3(true, x, y) → gcd(x, minus(y, x))
if3(false, x, y) → y
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gcd(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(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:

IF(true, x, y) → MINUS(x, y)
MINUS(s(x), y) → IF(gt(s(x), y), x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gcd(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(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].

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gcd(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

(25) Obligation:

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

IF(true, x, y) → MINUS(x, y)
MINUS(s(x), y) → IF(gt(s(x), y), x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))

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

(26) QDPSizeChangeProof (EQUIVALENT transformation)

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

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

MINUS(s(x), y) → IF(gt(s(x), y), x, y)
The graph contains the following edges 1 > 2, 2 >= 3
IF(true, x, y) → MINUS(x, y)
The graph contains the following edges 2 >= 1, 3 >= 2

(27) YES

(28) Obligation:

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

IF2(true, x, y) → GCD(minus(x, y), y)
GCD(x, y) → IF1(ge(x, y), x, y)
IF1(true, x, y) → IF2(gt(y, 0), x, y)
IF1(false, x, y) → IF3(gt(x, 0), x, y)
IF3(true, x, y) → GCD(x, minus(y, x))

The TRS R consists of the following rules:

minus(s(x), y) → if(gt(s(x), y), x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gcd(x, y) → if1(ge(x, y), x, y)
if1(true, x, y) → if2(gt(y, 0), x, y)
if1(false, x, y) → if3(gt(x, 0), x, y)
if2(true, x, y) → gcd(minus(x, y), y)
if2(false, x, y) → x
if3(true, x, y) → gcd(x, minus(y, x))
if3(false, x, y) → y
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gcd(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(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:

IF2(true, x, y) → GCD(minus(x, y), y)
GCD(x, y) → IF1(ge(x, y), x, y)
IF1(true, x, y) → IF2(gt(y, 0), x, y)
IF1(false, x, y) → IF3(gt(x, 0), x, y)
IF3(true, x, y) → GCD(x, minus(y, x))

The TRS R consists of the following rules:

minus(s(x), y) → if(gt(s(x), y), x, y)
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gt(0, y) → false
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gcd(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(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].

gcd(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)
if3(true, x0, x1)
if3(false, x0, x1)

(32) Obligation:

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

IF2(true, x, y) → GCD(minus(x, y), y)
GCD(x, y) → IF1(ge(x, y), x, y)
IF1(true, x, y) → IF2(gt(y, 0), x, y)
IF1(false, x, y) → IF3(gt(x, 0), x, y)
IF3(true, x, y) → GCD(x, minus(y, x))

The TRS R consists of the following rules:

minus(s(x), y) → if(gt(s(x), y), x, y)
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gt(0, y) → false
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(33) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule GCD(x, y) → IF1(ge(x, y), x, y) at position [0] we obtained the following new rules [LPAR04]:

GCD(x0, 0) → IF1(true, x0, 0) → GCD(x0, 0) → IF1(true, x0, 0)
GCD(0, s(x0)) → IF1(false, 0, s(x0)) → GCD(0, s(x0)) → IF1(false, 0, s(x0))
GCD(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1)) → GCD(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1))

(34) Obligation:

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

IF2(true, x, y) → GCD(minus(x, y), y)
IF1(true, x, y) → IF2(gt(y, 0), x, y)
IF1(false, x, y) → IF3(gt(x, 0), x, y)
IF3(true, x, y) → GCD(x, minus(y, x))
GCD(x0, 0) → IF1(true, x0, 0)
GCD(0, s(x0)) → IF1(false, 0, s(x0))
GCD(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

minus(s(x), y) → if(gt(s(x), y), x, y)
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gt(0, y) → false
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(35) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule IF1(true, x, y) → IF2(gt(y, 0), x, y) at position [0] we obtained the following new rules [LPAR04]:

IF1(true, y0, s(x0)) → IF2(true, y0, s(x0)) → IF1(true, y0, s(x0)) → IF2(true, y0, s(x0))
IF1(true, y0, 0) → IF2(false, y0, 0) → IF1(true, y0, 0) → IF2(false, y0, 0)

(36) Obligation:

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

IF2(true, x, y) → GCD(minus(x, y), y)
IF1(false, x, y) → IF3(gt(x, 0), x, y)
IF3(true, x, y) → GCD(x, minus(y, x))
GCD(x0, 0) → IF1(true, x0, 0)
GCD(0, s(x0)) → IF1(false, 0, s(x0))
GCD(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1))
IF1(true, y0, s(x0)) → IF2(true, y0, s(x0))
IF1(true, y0, 0) → IF2(false, y0, 0)

The TRS R consists of the following rules:

minus(s(x), y) → if(gt(s(x), y), x, y)
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gt(0, y) → false
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(37) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(38) Obligation:

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

GCD(0, s(x0)) → IF1(false, 0, s(x0))
IF1(false, x, y) → IF3(gt(x, 0), x, y)
IF3(true, x, y) → GCD(x, minus(y, x))
GCD(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1))
IF1(true, y0, s(x0)) → IF2(true, y0, s(x0))
IF2(true, x, y) → GCD(minus(x, y), y)

The TRS R consists of the following rules:

minus(s(x), y) → if(gt(s(x), y), x, y)
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gt(0, y) → false
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(39) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule IF1(false, x, y) → IF3(gt(x, 0), x, y) at position [0] we obtained the following new rules [LPAR04]:

IF1(false, s(x0), y1) → IF3(true, s(x0), y1) → IF1(false, s(x0), y1) → IF3(true, s(x0), y1)
IF1(false, 0, y1) → IF3(false, 0, y1) → IF1(false, 0, y1) → IF3(false, 0, y1)

(40) Obligation:

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

GCD(0, s(x0)) → IF1(false, 0, s(x0))
IF3(true, x, y) → GCD(x, minus(y, x))
GCD(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1))
IF1(true, y0, s(x0)) → IF2(true, y0, s(x0))
IF2(true, x, y) → GCD(minus(x, y), y)
IF1(false, s(x0), y1) → IF3(true, s(x0), y1)
IF1(false, 0, y1) → IF3(false, 0, y1)

The TRS R consists of the following rules:

minus(s(x), y) → if(gt(s(x), y), x, y)
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gt(0, y) → false
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(41) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(42) Obligation:

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

GCD(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1))
IF1(true, y0, s(x0)) → IF2(true, y0, s(x0))
IF2(true, x, y) → GCD(minus(x, y), y)
IF1(false, s(x0), y1) → IF3(true, s(x0), y1)
IF3(true, x, y) → GCD(x, minus(y, x))

The TRS R consists of the following rules:

minus(s(x), y) → if(gt(s(x), y), x, y)
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gt(0, y) → false
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(43) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule IF2(true, x, y) → GCD(minus(x, y), y) at position [0] we obtained the following new rules [LPAR04]:

IF2(true, s(x0), x1) → GCD(if(gt(s(x0), x1), x0, x1), x1) → IF2(true, s(x0), x1) → GCD(if(gt(s(x0), x1), x0, x1), x1)

(44) Obligation:

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

GCD(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1))
IF1(true, y0, s(x0)) → IF2(true, y0, s(x0))
IF1(false, s(x0), y1) → IF3(true, s(x0), y1)
IF3(true, x, y) → GCD(x, minus(y, x))
IF2(true, s(x0), x1) → GCD(if(gt(s(x0), x1), x0, x1), x1)

The TRS R consists of the following rules:

minus(s(x), y) → if(gt(s(x), y), x, y)
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gt(0, y) → false
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(45) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule IF3(true, x, y) → GCD(x, minus(y, x)) at position [1] we obtained the following new rules [LPAR04]:

IF3(true, x1, s(x0)) → GCD(x1, if(gt(s(x0), x1), x0, x1)) → IF3(true, x1, s(x0)) → GCD(x1, if(gt(s(x0), x1), x0, x1))

(46) Obligation:

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

GCD(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1))
IF1(true, y0, s(x0)) → IF2(true, y0, s(x0))
IF1(false, s(x0), y1) → IF3(true, s(x0), y1)
IF2(true, s(x0), x1) → GCD(if(gt(s(x0), x1), x0, x1), x1)
IF3(true, x1, s(x0)) → GCD(x1, if(gt(s(x0), x1), x0, x1))

The TRS R consists of the following rules:

minus(s(x), y) → if(gt(s(x), y), x, y)
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gt(0, y) → false
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(47) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF1(true, y0, s(x0)) → IF2(true, y0, s(x0)) we obtained the following new rules [LPAR04]:

IF1(true, s(z0), s(z1)) → IF2(true, s(z0), s(z1)) → IF1(true, s(z0), s(z1)) → IF2(true, s(z0), s(z1))

(48) Obligation:

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

GCD(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1))
IF1(false, s(x0), y1) → IF3(true, s(x0), y1)
IF2(true, s(x0), x1) → GCD(if(gt(s(x0), x1), x0, x1), x1)
IF3(true, x1, s(x0)) → GCD(x1, if(gt(s(x0), x1), x0, x1))
IF1(true, s(z0), s(z1)) → IF2(true, s(z0), s(z1))

The TRS R consists of the following rules:

minus(s(x), y) → if(gt(s(x), y), x, y)
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gt(0, y) → false
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(49) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF1(false, s(x0), y1) → IF3(true, s(x0), y1) we obtained the following new rules [LPAR04]:

IF1(false, s(z0), s(z1)) → IF3(true, s(z0), s(z1)) → IF1(false, s(z0), s(z1)) → IF3(true, s(z0), s(z1))

(50) Obligation:

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

GCD(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1))
IF2(true, s(x0), x1) → GCD(if(gt(s(x0), x1), x0, x1), x1)
IF3(true, x1, s(x0)) → GCD(x1, if(gt(s(x0), x1), x0, x1))
IF1(true, s(z0), s(z1)) → IF2(true, s(z0), s(z1))
IF1(false, s(z0), s(z1)) → IF3(true, s(z0), s(z1))

The TRS R consists of the following rules:

minus(s(x), y) → if(gt(s(x), y), x, y)
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gt(0, y) → false
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(51) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF2(true, s(x0), x1) → GCD(if(gt(s(x0), x1), x0, x1), x1) we obtained the following new rules [LPAR04]:

IF2(true, s(z0), s(z1)) → GCD(if(gt(s(z0), s(z1)), z0, s(z1)), s(z1)) → IF2(true, s(z0), s(z1)) → GCD(if(gt(s(z0), s(z1)), z0, s(z1)), s(z1))

(52) Obligation:

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

GCD(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1))
IF3(true, x1, s(x0)) → GCD(x1, if(gt(s(x0), x1), x0, x1))
IF1(true, s(z0), s(z1)) → IF2(true, s(z0), s(z1))
IF1(false, s(z0), s(z1)) → IF3(true, s(z0), s(z1))
IF2(true, s(z0), s(z1)) → GCD(if(gt(s(z0), s(z1)), z0, s(z1)), s(z1))

The TRS R consists of the following rules:

minus(s(x), y) → if(gt(s(x), y), x, y)
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gt(0, y) → false
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(53) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IF2(true, s(z0), s(z1)) → GCD(if(gt(s(z0), s(z1)), z0, s(z1)), s(z1)) at position [0,0] we obtained the following new rules [LPAR04]:

IF2(true, s(z0), s(z1)) → GCD(if(gt(z0, z1), z0, s(z1)), s(z1)) → IF2(true, s(z0), s(z1)) → GCD(if(gt(z0, z1), z0, s(z1)), s(z1))

(54) Obligation:

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

GCD(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1))
IF3(true, x1, s(x0)) → GCD(x1, if(gt(s(x0), x1), x0, x1))
IF1(true, s(z0), s(z1)) → IF2(true, s(z0), s(z1))
IF1(false, s(z0), s(z1)) → IF3(true, s(z0), s(z1))
IF2(true, s(z0), s(z1)) → GCD(if(gt(z0, z1), z0, s(z1)), s(z1))

The TRS R consists of the following rules:

minus(s(x), y) → if(gt(s(x), y), x, y)
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gt(0, y) → false
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(55) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF3(true, x1, s(x0)) → GCD(x1, if(gt(s(x0), x1), x0, x1)) we obtained the following new rules [LPAR04]:

IF3(true, s(z0), s(z1)) → GCD(s(z0), if(gt(s(z1), s(z0)), z1, s(z0))) → IF3(true, s(z0), s(z1)) → GCD(s(z0), if(gt(s(z1), s(z0)), z1, s(z0)))

(56) Obligation:

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

GCD(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1))
IF1(true, s(z0), s(z1)) → IF2(true, s(z0), s(z1))
IF1(false, s(z0), s(z1)) → IF3(true, s(z0), s(z1))
IF2(true, s(z0), s(z1)) → GCD(if(gt(z0, z1), z0, s(z1)), s(z1))
IF3(true, s(z0), s(z1)) → GCD(s(z0), if(gt(s(z1), s(z0)), z1, s(z0)))

The TRS R consists of the following rules:

minus(s(x), y) → if(gt(s(x), y), x, y)
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gt(0, y) → false
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(57) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IF3(true, s(z0), s(z1)) → GCD(s(z0), if(gt(s(z1), s(z0)), z1, s(z0))) at position [1,0] we obtained the following new rules [LPAR04]:

IF3(true, s(z0), s(z1)) → GCD(s(z0), if(gt(z1, z0), z1, s(z0))) → IF3(true, s(z0), s(z1)) → GCD(s(z0), if(gt(z1, z0), z1, s(z0)))

(58) Obligation:

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

GCD(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1))
IF1(true, s(z0), s(z1)) → IF2(true, s(z0), s(z1))
IF1(false, s(z0), s(z1)) → IF3(true, s(z0), s(z1))
IF2(true, s(z0), s(z1)) → GCD(if(gt(z0, z1), z0, s(z1)), s(z1))
IF3(true, s(z0), s(z1)) → GCD(s(z0), if(gt(z1, z0), z1, s(z0)))

The TRS R consists of the following rules:

minus(s(x), y) → if(gt(s(x), y), x, y)
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gt(0, y) → false
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(59) Induction-Processor (SOUND transformation)

This DP could be deleted by the Induction-Processor:
IF2(true_renamed, s(z0''), s(z1'')) → GCD(if(gt(z0'', z1''), z0'', s(z1'')), s(z1''))

This order was computed:
Polynomial interpretation [POLO]:

POL(0) = 0
POL(GCD(x₁, x₂)) = 1 + x₁ + x₂
POL(IF1(x₁, x₂, x₃)) = 1 + x₂ + x₃
POL(IF2(x₁, x₂, x₃)) = 1 + x₂ + x₃
POL(IF3(x₁, x₂, x₃)) = 1 + x₂ + x₃
POL(false_renamed) = 0
POL(ge(x₁, x₂)) = x₂
POL(gt(x₁, x₂)) = 0
POL(if(x₁, x₂, x₃)) = 1 + x₁ + x₂
POL(minus(x₁, x₂)) = x₁
POL(s(x₁)) = 1 + x₁
POL(true_renamed) = 0

At least one of these decreasing rules is always used after the deleted DP:
if(false_renamed, x2, y'') → 0

The following formula is valid:
z0'':sort[a0],z1'':sort[a0].if'(gt(z0'' , z1'' ), z0'' , s(z1'' ))=true

The transformed set:
if'(true_renamed, x'', y') → minus'(x'', y')
if'(false_renamed, x2, y'') → true
minus'(s(x6), y3) → if'(gt(s(x6), y3), x6, y3)
minus'(0, v58) → false
gt(s(x), 0) → true_renamed
gt(s(x'), s(y)) → gt(x', y)
if(true_renamed, x'', y') → s(minus(x'', y'))
if(false_renamed, x2, y'') → 0
gt(0, y1) → false_renamed
ge(x3, 0) → true_renamed
ge(0, s(x4)) → false_renamed
ge(s(x5), s(y2)) → ge(x5, y2)
minus(s(x6), y3) → if(gt(s(x6), y3), x6, y3)
minus(0, v58) → 0
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](s(v60), s(v61)) → equal_sort[a0](v60, v61)
equal_sort[a0](s(v60), 0) → false
equal_sort[a0](0, s(v62)) → false
equal_sort[a0](0, 0) → true
equal_sort[a26](true_renamed, true_renamed) → true
equal_sort[a26](true_renamed, false_renamed) → false
equal_sort[a26](false_renamed, true_renamed) → false
equal_sort[a26](false_renamed, false_renamed) → true
equal_sort[a44](witness_sort[a44], witness_sort[a44]) → 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, v60, v61, v62, x2, y'', x6, y', v58, y3, x', y, y1, x3, x4, x5, y2]
   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](s(v60), s(v61)) -> equal_sort[a0](v60, v61)
   equal_sort[a0](s(v60), 0) -> false
   equal_sort[a0](0, s(v62)) -> false
   equal_sort[a0](0, 0) -> true
   equal_sort[a26](true_renamed, true_renamed) -> true
   equal_sort[a26](true_renamed, false_renamed) -> false
   equal_sort[a26](false_renamed, true_renamed) -> false
   equal_sort[a26](false_renamed, false_renamed) -> true
   equal_sort[a44](witness_sort[a44], witness_sort[a44]) -> true
   if'(false_renamed, x2, y'') -> true
   if'(true_renamed, s(x6), y') -> if'(gt(s(x6), y'), x6, y')
   if'(true_renamed, 0, y') -> false
   minus'(0, v58) -> false
   equal_sort[a26](gt(s(x6), y3), true_renamed) -> true | minus'(s(x6), y3) -> minus'(x6, y3)
   equal_sort[a26](gt(s(x6), y3), true_renamed) -> false | minus'(s(x6), y3) -> true
   gt(s(x), 0) -> true_renamed
   gt(s(x'), s(y)) -> gt(x', y)
   gt(0, y1) -> false_renamed
   if(false_renamed, x2, y'') -> 0
   if(true_renamed, s(x6), y') -> s(if(gt(s(x6), y'), x6, y'))
   if(true_renamed, 0, y') -> s(0)
   ge(x3, 0) -> true_renamed
   ge(0, s(x4)) -> false_renamed
   ge(s(x5), s(y2)) -> ge(x5, y2)
   minus(0, v58) -> 0
   equal_sort[a26](gt(s(x6), y3), true_renamed) -> true | minus(s(x6), y3) -> s(minus(x6, y3))
   equal_sort[a26](gt(s(x6), y3), true_renamed) -> false | minus(s(x6), y3) -> 0

using the following formula:
z0'':sort[a0],z1'':sort[a0].if'(gt(z0'', z1''), z0'', s(z1''))=true

could be successfully shown:
(0) Formula
(1) Induction by algorithm [EQUIVALENT, 0 ms]
(2) AND
    (3) Formula
        (4) Symbolic evaluation [EQUIVALENT, 0 ms]
        (5) Formula
        (6) Induction by data structure [EQUIVALENT, 0 ms]
        (7) AND
            (8) Formula
                (9) Symbolic evaluation under hypothesis [EQUIVALENT, 0 ms]
                (10) YES
            (11) Formula
                (12) Symbolic evaluation [EQUIVALENT, 0 ms]
                (13) YES
    (14) Formula
        (15) Symbolic evaluation [EQUIVALENT, 0 ms]
        (16) YES
    (17) Formula
        (18) Symbolic evaluation [EQUIVALENT, 0 ms]
        (19) Formula
        (20) Hypothesis Lifting [EQUIVALENT, 0 ms]
        (21) Formula
        (22) Inverse Substitution [SOUND, 0 ms]
        (23) Formula
        (24) Inverse Substitution [SOUND, 0 ms]
        (25) Formula
        (26) Induction by algorithm [EQUIVALENT, 0 ms]
        (27) AND
            (28) Formula
                (29) Symbolic evaluation [EQUIVALENT, 0 ms]
                (30) YES
            (31) Formula
                (32) Symbolic evaluation [EQUIVALENT, 0 ms]
                (33) YES
            (34) Formula
                (35) Symbolic evaluation under hypothesis [EQUIVALENT, 0 ms]
                (36) YES


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

(0)
Obligation:
Formula:
z0'':sort[a0],z1'':sort[a0].if'(gt(z0'', z1''), z0'', s(z1''))=true

There are no hypotheses.




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

(1) Induction by algorithm (EQUIVALENT)
Induction by algorithm gt(z0'', z1'') generates the following cases:

1. Base Case:
Formula:
x:sort[a0].if'(gt(s(x), 0), s(x), s(0))=true

There are no hypotheses.





2. Base Case:
Formula:
y1:sort[a0].if'(gt(0, y1), 0, s(y1))=true

There are no hypotheses.





1. Step Case:
Formula:
x':sort[a0],y:sort[a0].if'(gt(s(x'), s(y)), s(x'), s(s(y)))=true

Hypotheses:
x':sort[a0],y:sort[a0].if'(gt(x', y), x', s(y))=true






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

(2)
Complex Obligation (AND)

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

(3)
Obligation:
Formula:
x:sort[a0].if'(gt(s(x), 0), s(x), s(0))=true

There are no hypotheses.




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

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

(5)
Obligation:
Formula:
x:sort[a0].if'(gt(x, 0), x, s(0))=true

There are no hypotheses.




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

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



1. Base Case:
Formula:
if'(gt(0, 0), 0, s(0))=true

There are no hypotheses.





1. Step Case:
Formula:
n:sort[a0].if'(gt(s(n), 0), s(n), s(0))=true

Hypotheses:
n:sort[a0].if'(gt(n, 0), n, s(0))=true






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

(7)
Complex Obligation (AND)

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

(8)
Obligation:
Formula:
n:sort[a0].if'(gt(s(n), 0), s(n), s(0))=true

Hypotheses:
n:sort[a0].if'(gt(n, 0), n, s(0))=true




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

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

n:sort[a0].if'(gt(n, 0), n, s(0))=true

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

(10)
YES

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

(11)
Obligation:
Formula:
if'(gt(0, 0), 0, s(0))=true

There are no hypotheses.




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

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

(13)
YES

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

(14)
Obligation:
Formula:
y1:sort[a0].if'(gt(0, y1), 0, s(y1))=true

There are no hypotheses.




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

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

(16)
YES

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

(17)
Obligation:
Formula:
x':sort[a0],y:sort[a0].if'(gt(s(x'), s(y)), s(x'), s(s(y)))=true

Hypotheses:
x':sort[a0],y:sort[a0].if'(gt(x', y), x', s(y))=true




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

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

(19)
Obligation:
Formula:
x':sort[a0],y:sort[a0].if'(gt(x', y), s(x'), s(s(y)))=true

Hypotheses:
x':sort[a0],y:sort[a0].if'(gt(x', y), x', s(y))=true




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

(20) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
x':sort[a0],y:sort[a0].(if'(gt(x', y), x', s(y))=true->if'(gt(x', y), s(x'), s(s(y)))=true)

There are no hypotheses.




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

(21)
Obligation:
Formula:
x':sort[a0],y:sort[a0].(if'(gt(x', y), x', s(y))=true->if'(gt(x', y), s(x'), s(s(y)))=true)

There are no hypotheses.




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

(22) Inverse Substitution (SOUND)
The formula could be generalised by inverse substitution to:
n:sort[a26],x':sort[a0],y:sort[a0].(if'(n, x', s(y))=true->if'(n, s(x'), s(s(y)))=true)

Inverse substitution used:
[gt(x', y)/n]


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

(23)
Obligation:
Formula:
n:sort[a26],x':sort[a0],y:sort[a0].(if'(n, x', s(y))=true->if'(n, s(x'), s(s(y)))=true)

There are no hypotheses.




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

(24) Inverse Substitution (SOUND)
The formula could be generalised by inverse substitution to:
n:sort[a26],x':sort[a0],n':sort[a0].(if'(n, x', n')=true->if'(n, s(x'), s(n'))=true)

Inverse substitution used:
[s(y)/n']


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

(25)
Obligation:
Formula:
n:sort[a26],x':sort[a0],n':sort[a0].(if'(n, x', n')=true->if'(n, s(x'), s(n'))=true)

There are no hypotheses.




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

(26) Induction by algorithm (EQUIVALENT)
Induction by algorithm if'(n, x', n') generates the following cases:

1. Base Case:
Formula:
x2:sort[a0],y'':sort[a0].(if'(false_renamed, x2, y'')=true->if'(false_renamed, s(x2), s(y''))=true)

There are no hypotheses.





2. Base Case:
Formula:
y':sort[a0].(if'(true_renamed, 0, y')=true->if'(true_renamed, s(0), s(y'))=true)

There are no hypotheses.





1. Step Case:
Formula:
x6:sort[a0],y':sort[a0].(if'(true_renamed, s(x6), y')=true->if'(true_renamed, s(s(x6)), s(y'))=true)

Hypotheses:
x6:sort[a0],y':sort[a0].(if'(gt(s(x6), y'), x6, y')=true->if'(gt(s(x6), y'), s(x6), s(y'))=true)






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

(27)
Complex Obligation (AND)

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

(28)
Obligation:
Formula:
x2:sort[a0],y'':sort[a0].(if'(false_renamed, x2, y'')=true->if'(false_renamed, s(x2), s(y''))=true)

There are no hypotheses.




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

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

(30)
YES

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

(31)
Obligation:
Formula:
y':sort[a0].(if'(true_renamed, 0, y')=true->if'(true_renamed, s(0), s(y'))=true)

There are no hypotheses.




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

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

(33)
YES

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

(34)
Obligation:
Formula:
x6:sort[a0],y':sort[a0].(if'(true_renamed, s(x6), y')=true->if'(true_renamed, s(s(x6)), s(y'))=true)

Hypotheses:
x6:sort[a0],y':sort[a0].(if'(gt(s(x6), y'), x6, y')=true->if'(gt(s(x6), y'), s(x6), s(y'))=true)




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

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

x6:sort[a0],y':sort[a0].(if'(gt(s(x6), y'), x6, y')=true->if'(gt(s(x6), y'), s(x6), s(y'))=true)

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

(36)
YES

(60) Complex Obligation (AND)

(61) Obligation:

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

GCD(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1))
IF1(true, s(z0), s(z1)) → IF2(true, s(z0), s(z1))
IF1(false, s(z0), s(z1)) → IF3(true, s(z0), s(z1))
IF3(true, s(z0), s(z1)) → GCD(s(z0), if(gt(z1, z0), z1, s(z0)))

The TRS R consists of the following rules:

minus(s(x), y) → if(gt(s(x), y), x, y)
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gt(0, y) → false
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(62) DependencyGraphProof (EQUIVALENT transformation)

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

(63) Obligation:

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

IF1(false, s(z0), s(z1)) → IF3(true, s(z0), s(z1))
IF3(true, s(z0), s(z1)) → GCD(s(z0), if(gt(z1, z0), z1, s(z0)))
GCD(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

minus(s(x), y) → if(gt(s(x), y), x, y)
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gt(0, y) → false
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(64) Induction-Processor (SOUND transformation)

This DP could be deleted by the Induction-Processor:
IF3(true_renamed, s(z0'), s(z1')) → GCD(s(z0'), if(gt(z1', z0'), z1', s(z0')))

This order was computed:
Polynomial interpretation [POLO]:

POL(0) = 0
POL(GCD(x₁, x₂)) = x₂
POL(IF1(x₁, x₂, x₃)) = x₃
POL(IF3(x₁, x₂, x₃)) = x₃
POL(false_renamed) = 1
POL(ge(x₁, x₂)) = x₁ + x₂
POL(gt(x₁, x₂)) = 1 + x₁ + x₂
POL(if(x₁, x₂, x₃)) = 1 + x₂
POL(minus(x₁, x₂)) = x₁
POL(s(x₁)) = 1 + x₁
POL(true_renamed) = 0

At least one of these decreasing rules is always used after the deleted DP:
if(false_renamed, x2, y'') → 0

The following formula is valid:
z1':sort[a0],z0':sort[a0].if'(gt(z1' , z0' ), z1' , s(z0' ))=true

The transformed set:
if'(true_renamed, x'', y') → minus'(x'', y')
if'(false_renamed, x2, y'') → true
minus'(s(x6), y3) → if'(gt(s(x6), y3), x6, y3)
minus'(0, v28) → false
gt(s(x), 0) → true_renamed
gt(s(x'), s(y)) → gt(x', y)
if(true_renamed, x'', y') → s(minus(x'', y'))
if(false_renamed, x2, y'') → 0
gt(0, y1) → false_renamed
ge(x3, 0) → true_renamed
ge(0, s(x4)) → false_renamed
ge(s(x5), s(y2)) → ge(x5, y2)
minus(s(x6), y3) → if(gt(s(x6), y3), x6, y3)
minus(0, v28) → 0
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](s(v30), s(v31)) → equal_sort[a0](v30, v31)
equal_sort[a0](s(v30), 0) → false
equal_sort[a0](0, s(v32)) → false
equal_sort[a0](0, 0) → true
equal_sort[a22](true_renamed, true_renamed) → true
equal_sort[a22](true_renamed, false_renamed) → false
equal_sort[a22](false_renamed, true_renamed) → false
equal_sort[a22](false_renamed, false_renamed) → true
equal_sort[a41](witness_sort[a41], witness_sort[a41]) → 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, v30, v31, v32, x2, y'', x6, y', v28, y3, x', y, y1, x3, x4, x5, y2]
   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](s(v30), s(v31)) -> equal_sort[a0](v30, v31)
   equal_sort[a0](s(v30), 0) -> false
   equal_sort[a0](0, s(v32)) -> false
   equal_sort[a0](0, 0) -> true
   equal_sort[a22](true_renamed, true_renamed) -> true
   equal_sort[a22](true_renamed, false_renamed) -> false
   equal_sort[a22](false_renamed, true_renamed) -> false
   equal_sort[a22](false_renamed, false_renamed) -> true
   equal_sort[a41](witness_sort[a41], witness_sort[a41]) -> true
   if'(false_renamed, x2, y'') -> true
   if'(true_renamed, s(x6), y') -> if'(gt(s(x6), y'), x6, y')
   if'(true_renamed, 0, y') -> false
   minus'(0, v28) -> false
   equal_sort[a22](gt(s(x6), y3), true_renamed) -> true | minus'(s(x6), y3) -> minus'(x6, y3)
   equal_sort[a22](gt(s(x6), y3), true_renamed) -> false | minus'(s(x6), y3) -> true
   gt(s(x), 0) -> true_renamed
   gt(s(x'), s(y)) -> gt(x', y)
   gt(0, y1) -> false_renamed
   if(false_renamed, x2, y'') -> 0
   if(true_renamed, s(x6), y') -> s(if(gt(s(x6), y'), x6, y'))
   if(true_renamed, 0, y') -> s(0)
   ge(x3, 0) -> true_renamed
   ge(0, s(x4)) -> false_renamed
   ge(s(x5), s(y2)) -> ge(x5, y2)
   minus(0, v28) -> 0
   equal_sort[a22](gt(s(x6), y3), true_renamed) -> true | minus(s(x6), y3) -> s(minus(x6, y3))
   equal_sort[a22](gt(s(x6), y3), true_renamed) -> false | minus(s(x6), y3) -> 0

using the following formula:
z1':sort[a0],z0':sort[a0].if'(gt(z1', z0'), z1', s(z0'))=true

could be successfully shown:
(0) Formula
(1) Induction by algorithm [EQUIVALENT, 0 ms]
(2) AND
    (3) Formula
        (4) Symbolic evaluation [EQUIVALENT, 0 ms]
        (5) Formula
        (6) Induction by data structure [EQUIVALENT, 0 ms]
        (7) AND
            (8) Formula
                (9) Symbolic evaluation under hypothesis [EQUIVALENT, 0 ms]
                (10) YES
            (11) Formula
                (12) Symbolic evaluation [EQUIVALENT, 0 ms]
                (13) YES
    (14) Formula
        (15) Symbolic evaluation [EQUIVALENT, 0 ms]
        (16) YES
    (17) Formula
        (18) Symbolic evaluation [EQUIVALENT, 0 ms]
        (19) Formula
        (20) Hypothesis Lifting [EQUIVALENT, 0 ms]
        (21) Formula
        (22) Inverse Substitution [SOUND, 0 ms]
        (23) Formula
        (24) Inverse Substitution [SOUND, 0 ms]
        (25) Formula
        (26) Induction by algorithm [EQUIVALENT, 0 ms]
        (27) AND
            (28) Formula
                (29) Symbolic evaluation [EQUIVALENT, 0 ms]
                (30) YES
            (31) Formula
                (32) Symbolic evaluation [EQUIVALENT, 0 ms]
                (33) YES
            (34) Formula
                (35) Symbolic evaluation under hypothesis [EQUIVALENT, 0 ms]
                (36) YES


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

(0)
Obligation:
Formula:
z1':sort[a0],z0':sort[a0].if'(gt(z1', z0'), z1', s(z0'))=true

There are no hypotheses.




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

(1) Induction by algorithm (EQUIVALENT)
Induction by algorithm gt(z1', z0') generates the following cases:

1. Base Case:
Formula:
x:sort[a0].if'(gt(s(x), 0), s(x), s(0))=true

There are no hypotheses.





2. Base Case:
Formula:
y1:sort[a0].if'(gt(0, y1), 0, s(y1))=true

There are no hypotheses.





1. Step Case:
Formula:
x':sort[a0],y:sort[a0].if'(gt(s(x'), s(y)), s(x'), s(s(y)))=true

Hypotheses:
x':sort[a0],y:sort[a0].if'(gt(x', y), x', s(y))=true






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

(2)
Complex Obligation (AND)

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

(3)
Obligation:
Formula:
x:sort[a0].if'(gt(s(x), 0), s(x), s(0))=true

There are no hypotheses.




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

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

(5)
Obligation:
Formula:
x:sort[a0].if'(gt(x, 0), x, s(0))=true

There are no hypotheses.




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

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



1. Base Case:
Formula:
if'(gt(0, 0), 0, s(0))=true

There are no hypotheses.





1. Step Case:
Formula:
n:sort[a0].if'(gt(s(n), 0), s(n), s(0))=true

Hypotheses:
n:sort[a0].if'(gt(n, 0), n, s(0))=true






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

(7)
Complex Obligation (AND)

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

(8)
Obligation:
Formula:
n:sort[a0].if'(gt(s(n), 0), s(n), s(0))=true

Hypotheses:
n:sort[a0].if'(gt(n, 0), n, s(0))=true




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

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

n:sort[a0].if'(gt(n, 0), n, s(0))=true

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

(10)
YES

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

(11)
Obligation:
Formula:
if'(gt(0, 0), 0, s(0))=true

There are no hypotheses.




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

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

(13)
YES

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

(14)
Obligation:
Formula:
y1:sort[a0].if'(gt(0, y1), 0, s(y1))=true

There are no hypotheses.




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

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

(16)
YES

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

(17)
Obligation:
Formula:
x':sort[a0],y:sort[a0].if'(gt(s(x'), s(y)), s(x'), s(s(y)))=true

Hypotheses:
x':sort[a0],y:sort[a0].if'(gt(x', y), x', s(y))=true




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

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

(19)
Obligation:
Formula:
x':sort[a0],y:sort[a0].if'(gt(x', y), s(x'), s(s(y)))=true

Hypotheses:
x':sort[a0],y:sort[a0].if'(gt(x', y), x', s(y))=true




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

(20) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
x':sort[a0],y:sort[a0].(if'(gt(x', y), x', s(y))=true->if'(gt(x', y), s(x'), s(s(y)))=true)

There are no hypotheses.




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

(21)
Obligation:
Formula:
x':sort[a0],y:sort[a0].(if'(gt(x', y), x', s(y))=true->if'(gt(x', y), s(x'), s(s(y)))=true)

There are no hypotheses.




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

(22) Inverse Substitution (SOUND)
The formula could be generalised by inverse substitution to:
n:sort[a22],x':sort[a0],y:sort[a0].(if'(n, x', s(y))=true->if'(n, s(x'), s(s(y)))=true)

Inverse substitution used:
[gt(x', y)/n]


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

(23)
Obligation:
Formula:
n:sort[a22],x':sort[a0],y:sort[a0].(if'(n, x', s(y))=true->if'(n, s(x'), s(s(y)))=true)

There are no hypotheses.




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

(24) Inverse Substitution (SOUND)
The formula could be generalised by inverse substitution to:
n:sort[a22],x':sort[a0],n':sort[a0].(if'(n, x', n')=true->if'(n, s(x'), s(n'))=true)

Inverse substitution used:
[s(y)/n']


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

(25)
Obligation:
Formula:
n:sort[a22],x':sort[a0],n':sort[a0].(if'(n, x', n')=true->if'(n, s(x'), s(n'))=true)

There are no hypotheses.




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

(26) Induction by algorithm (EQUIVALENT)
Induction by algorithm if'(n, x', n') generates the following cases:

1. Base Case:
Formula:
x2:sort[a0],y'':sort[a0].(if'(false_renamed, x2, y'')=true->if'(false_renamed, s(x2), s(y''))=true)

There are no hypotheses.





2. Base Case:
Formula:
y':sort[a0].(if'(true_renamed, 0, y')=true->if'(true_renamed, s(0), s(y'))=true)

There are no hypotheses.





1. Step Case:
Formula:
x6:sort[a0],y':sort[a0].(if'(true_renamed, s(x6), y')=true->if'(true_renamed, s(s(x6)), s(y'))=true)

Hypotheses:
x6:sort[a0],y':sort[a0].(if'(gt(s(x6), y'), x6, y')=true->if'(gt(s(x6), y'), s(x6), s(y'))=true)






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

(27)
Complex Obligation (AND)

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

(28)
Obligation:
Formula:
x2:sort[a0],y'':sort[a0].(if'(false_renamed, x2, y'')=true->if'(false_renamed, s(x2), s(y''))=true)

There are no hypotheses.




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

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

(30)
YES

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

(31)
Obligation:
Formula:
y':sort[a0].(if'(true_renamed, 0, y')=true->if'(true_renamed, s(0), s(y'))=true)

There are no hypotheses.




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

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

(33)
YES

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

(34)
Obligation:
Formula:
x6:sort[a0],y':sort[a0].(if'(true_renamed, s(x6), y')=true->if'(true_renamed, s(s(x6)), s(y'))=true)

Hypotheses:
x6:sort[a0],y':sort[a0].(if'(gt(s(x6), y'), x6, y')=true->if'(gt(s(x6), y'), s(x6), s(y'))=true)




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

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

x6:sort[a0],y':sort[a0].(if'(gt(s(x6), y'), x6, y')=true->if'(gt(s(x6), y'), s(x6), s(y'))=true)

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

(36)
YES

(65) Complex Obligation (AND)

(66) Obligation:

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

IF1(false, s(z0), s(z1)) → IF3(true, s(z0), s(z1))
GCD(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

minus(s(x), y) → if(gt(s(x), y), x, y)
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
gt(0, y) → false
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(67) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(68) TRUE

(69) Obligation:

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

if'(true_renamed, x'', y') → minus'(x'', y')
if'(false_renamed, x2, y'') → true
minus'(s(x6), y3) → if'(gt(s(x6), y3), x6, y3)
minus'(0, v28) → false
gt(s(x), 0) → true_renamed
gt(s(x'), s(y)) → gt(x', y)
if(true_renamed, x'', y') → s(minus(x'', y'))
if(false_renamed, x2, y'') → 0
gt(0, y1) → false_renamed
ge(x3, 0) → true_renamed
ge(0, s(x4)) → false_renamed
ge(s(x5), s(y2)) → ge(x5, y2)
minus(s(x6), y3) → if(gt(s(x6), y3), x6, y3)
minus(0, v28) → 0
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](s(v30), s(v31)) → equal_sort[a0](v30, v31)
equal_sort[a0](s(v30), 0) → false
equal_sort[a0](0, s(v32)) → false
equal_sort[a0](0, 0) → true
equal_sort[a22](true_renamed, true_renamed) → true
equal_sort[a22](true_renamed, false_renamed) → false
equal_sort[a22](false_renamed, true_renamed) → false
equal_sort[a22](false_renamed, false_renamed) → true
equal_sort[a41](witness_sort[a41], witness_sort[a41]) → true

Q is empty.

(70) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Combined order from the following AFS and order.
if'(x1, x2, x3) = if'(x1, x2, x3)
true_renamed = true_renamed
minus'(x1, x2) = minus'(x1, x2)
false_renamed = false_renamed
true = true
s(x1) = s(x1)
gt(x1, x2) = gt(x1, x2)
0 = 0
false = false
if(x1, x2, x3) = if(x1, x2, x3)
minus(x1, x2) = minus(x1, x2)
ge(x1, x2) = ge(x1, x2)
equal_bool(x1, x2) = equal_bool(x1, x2)
and(x1, x2) = and(x1, x2)
or(x1, x2) = or(x1, x2)
not(x1) = not(x1)
isa_true(x1) = x1
isa_false(x1) = isa_false(x1)
equal_sort[a0](x1, x2) = equal_sort[a0](x1, x2)
equal_sort[a22](x1, x2) = equal_sort[a22](x1, x2)
equal_sort[a41](x1, x2) = equal_sort[a41](x1, x2)
witness_sort[a41] = witness_sort[a41]

Recursive path order with status [RPO].
Quasi-Precedence:

[if'₃, minus'₂] > [true_renamed, true, equal_bool₂, or₂] > s₁ > gt₂
[if'₃, minus'₂] > [true_renamed, true, equal_bool₂, or₂] > [false, and₂, equal_sort[a22]₂] > gt₂
[if₃, minus₂] > s₁ > gt₂
ge₂ > [false_renamed, 0] > [true_renamed, true, equal_bool₂, or₂] > s₁ > gt₂
ge₂ > [false_renamed, 0] > [true_renamed, true, equal_bool₂, or₂] > [false, and₂, equal_sort[a22]₂] > gt₂
not₁ > [true_renamed, true, equal_bool₂, or₂] > s₁ > gt₂
not₁ > [true_renamed, true, equal_bool₂, or₂] > [false, and₂, equal_sort[a22]₂] > gt₂
isa_false₁ > [true_renamed, true, equal_bool₂, or₂] > s₁ > gt₂
isa_false₁ > [true_renamed, true, equal_bool₂, or₂] > [false, and₂, equal_sort[a22]₂] > gt₂
equal_sort[a0]₂ > [true_renamed, true, equal_bool₂, or₂] > s₁ > gt₂
equal_sort[a0]₂ > [true_renamed, true, equal_bool₂, or₂] > [false, and₂, equal_sort[a22]₂] > gt₂
equal_sort[a41]₂ > gt₂
witness_sort[a41] > [true_renamed, true, equal_bool₂, or₂] > s₁ > gt₂
witness_sort[a41] > [true_renamed, true, equal_bool₂, or₂] > [false, and₂, equal_sort[a22]₂] > gt₂

Status:

if'₃: [2,3,1]
true_renamed: multiset
minus'₂: [1,2]
false_renamed: multiset
true: multiset
s₁: multiset
gt₂: [1,2]
0: multiset
false: multiset
if₃: [2,3,1]
minus₂: [1,2]
ge₂: [1,2]
equal_bool₂: multiset
and₂: multiset
or₂: multiset
not₁: [1]
isa_false₁: multiset
equal_sort[a0]₂: multiset
equal_sort[a22]₂: multiset
equal_sort[a41]₂: [1,2]
witness_sort[a41]: multiset

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

if'(true_renamed, x'', y') → minus'(x'', y')
if'(false_renamed, x2, y'') → true
minus'(s(x6), y3) → if'(gt(s(x6), y3), x6, y3)
minus'(0, v28) → false
gt(s(x), 0) → true_renamed
gt(s(x'), s(y)) → gt(x', y)
if(true_renamed, x'', y') → s(minus(x'', y'))
if(false_renamed, x2, y'') → 0
gt(0, y1) → false_renamed
ge(x3, 0) → true_renamed
ge(0, s(x4)) → false_renamed
ge(s(x5), s(y2)) → ge(x5, y2)
minus(s(x6), y3) → if(gt(s(x6), y3), x6, y3)
minus(0, v28) → 0
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_false(true) → false
isa_false(false) → true
equal_sort[a0](s(v30), s(v31)) → equal_sort[a0](v30, v31)
equal_sort[a0](s(v30), 0) → false
equal_sort[a0](0, s(v32)) → false
equal_sort[a0](0, 0) → true
equal_sort[a22](true_renamed, true_renamed) → true
equal_sort[a22](true_renamed, false_renamed) → false
equal_sort[a22](false_renamed, true_renamed) → false
equal_sort[a22](false_renamed, false_renamed) → true
equal_sort[a41](witness_sort[a41], witness_sort[a41]) → true

(71) Obligation:

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

isa_true(true) → true
isa_true(false) → false

Q is empty.

(72) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Knuth-Bendix order [KBO] with precedence:

isa_true₁ > false > true

and weight map:

true=1
false=1
isa_true_1=0

The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

isa_true(true) → true
isa_true(false) → false

(73) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(74) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(75) YES

(76) Obligation:

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

if'(true_renamed, x'', y') → minus'(x'', y')
if'(false_renamed, x2, y'') → true
minus'(s(x6), y3) → if'(gt(s(x6), y3), x6, y3)
minus'(0, v58) → false
gt(s(x), 0) → true_renamed
gt(s(x'), s(y)) → gt(x', y)
if(true_renamed, x'', y') → s(minus(x'', y'))
if(false_renamed, x2, y'') → 0
gt(0, y1) → false_renamed
ge(x3, 0) → true_renamed
ge(0, s(x4)) → false_renamed
ge(s(x5), s(y2)) → ge(x5, y2)
minus(s(x6), y3) → if(gt(s(x6), y3), x6, y3)
minus(0, v58) → 0
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](s(v60), s(v61)) → equal_sort[a0](v60, v61)
equal_sort[a0](s(v60), 0) → false
equal_sort[a0](0, s(v62)) → false
equal_sort[a0](0, 0) → true
equal_sort[a26](true_renamed, true_renamed) → true
equal_sort[a26](true_renamed, false_renamed) → false
equal_sort[a26](false_renamed, true_renamed) → false
equal_sort[a26](false_renamed, false_renamed) → true
equal_sort[a44](witness_sort[a44], witness_sort[a44]) → true

Q is empty.

(77) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Combined order from the following AFS and order.
if'(x1, x2, x3) = if'(x1, x2, x3)
true_renamed = true_renamed
minus'(x1, x2) = minus'(x1, x2)
false_renamed = false_renamed
true = true
s(x1) = s(x1)
gt(x1, x2) = gt(x1, x2)
0 = 0
false = false
if(x1, x2, x3) = if(x1, x2, x3)
minus(x1, x2) = minus(x1, x2)
ge(x1, x2) = ge(x1, x2)
equal_bool(x1, x2) = equal_bool(x1, x2)
and(x1, x2) = and(x1, x2)
or(x1, x2) = or(x1, x2)
not(x1) = not(x1)
isa_true(x1) = x1
isa_false(x1) = isa_false(x1)
equal_sort[a0](x1, x2) = equal_sort[a0](x1, x2)
equal_sort[a26](x1, x2) = equal_sort[a26](x1, x2)
equal_sort[a44](x1, x2) = equal_sort[a44](x1, x2)
witness_sort[a44] = witness_sort[a44]

Recursive path order with status [RPO].
Quasi-Precedence:

[if₃, minus₂] > [0, false, not₁] > true_renamed > [if'₃, minus'₂, false_renamed, s₁] > true > gt₂
ge₂ > true_renamed > [if'₃, minus'₂, false_renamed, s₁] > true > gt₂
equal_bool₂ > true > gt₂
and₂ > gt₂
or₂ > true > gt₂
isa_false₁ > [0, false, not₁] > true_renamed > [if'₃, minus'₂, false_renamed, s₁] > true > gt₂
equal_sort[a0]₂ > [0, false, not₁] > true_renamed > [if'₃, minus'₂, false_renamed, s₁] > true > gt₂
equal_sort[a26]₂ > [0, false, not₁] > true_renamed > [if'₃, minus'₂, false_renamed, s₁] > true > gt₂
equal_sort[a44]₂ > gt₂
witness_sort[a44] > true > gt₂

Status:

if'₃: [3,2,1]
true_renamed: multiset
minus'₂: [2,1]
false_renamed: multiset
true: multiset
s₁: multiset
gt₂: [2,1]
0: multiset
false: multiset
if₃: [3,2,1]
minus₂: [2,1]
ge₂: [1,2]
equal_bool₂: multiset
and₂: multiset
or₂: multiset
not₁: multiset
isa_false₁: [1]
equal_sort[a0]₂: [1,2]
equal_sort[a26]₂: multiset
equal_sort[a44]₂: multiset
witness_sort[a44]: multiset

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

if'(true_renamed, x'', y') → minus'(x'', y')
if'(false_renamed, x2, y'') → true
minus'(s(x6), y3) → if'(gt(s(x6), y3), x6, y3)
minus'(0, v58) → false
gt(s(x), 0) → true_renamed
gt(s(x'), s(y)) → gt(x', y)
if(true_renamed, x'', y') → s(minus(x'', y'))
if(false_renamed, x2, y'') → 0
gt(0, y1) → false_renamed
ge(x3, 0) → true_renamed
ge(0, s(x4)) → false_renamed
ge(s(x5), s(y2)) → ge(x5, y2)
minus(s(x6), y3) → if(gt(s(x6), y3), x6, y3)
minus(0, v58) → 0
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_false(true) → false
isa_false(false) → true
equal_sort[a0](s(v60), s(v61)) → equal_sort[a0](v60, v61)
equal_sort[a0](s(v60), 0) → false
equal_sort[a0](0, s(v62)) → false
equal_sort[a0](0, 0) → true
equal_sort[a26](true_renamed, true_renamed) → true
equal_sort[a26](true_renamed, false_renamed) → false
equal_sort[a26](false_renamed, true_renamed) → false
equal_sort[a26](false_renamed, false_renamed) → true
equal_sort[a44](witness_sort[a44], witness_sort[a44]) → true

(78) Obligation:

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

isa_true(true) → true
isa_true(false) → false

Q is empty.

(79) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Knuth-Bendix order [KBO] with precedence:

isa_true₁ > false > true

and weight map:

true=1
false=1
isa_true_1=0

The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

isa_true(true) → true
isa_true(false) → false

(80) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(81) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.