/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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


YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) Overlay + Local Confluence [EQUIVALENT, 10 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QReductionProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (20) YES
    (21) QDP
        (22) UsableRulesProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) QReductionProof [EQUIVALENT, 0 ms]
        (25) QDP
        (26) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (27) YES
    (28) QDP
        (29) UsableRulesProof [EQUIVALENT, 0 ms]
        (30) QDP
        (31) QReductionProof [EQUIVALENT, 0 ms]
        (32) QDP
        (33) NonInfProof [EQUIVALENT, 46 ms]
        (34) QDP
        (35) TransformationProof [EQUIVALENT, 0 ms]
        (36) QDP
        (37) DependencyGraphProof [EQUIVALENT, 0 ms]
        (38) QDP
        (39) TransformationProof [EQUIVALENT, 0 ms]
        (40) QDP
        (41) DependencyGraphProof [EQUIVALENT, 0 ms]
        (42) QDP
        (43) TransformationProof [EQUIVALENT, 0 ms]
        (44) QDP
        (45) TransformationProof [EQUIVALENT, 0 ms]
        (46) QDP
        (47) TransformationProof [EQUIVALENT, 0 ms]
        (48) QDP
        (49) TransformationProof [EQUIVALENT, 0 ms]
        (50) QDP
        (51) Induction-Processor [SOUND, 46 ms]
        (52) AND
            (53) QDP
                (54) DependencyGraphProof [EQUIVALENT, 0 ms]
                (55) TRUE
            (56) QTRS
                (57) QTRSRRRProof [EQUIVALENT, 49 ms]
                (58) QTRS
                (59) RisEmptyProof [EQUIVALENT, 0 ms]
                (60) 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) -> gcd(y, x)
   if2(true, x, y) -> gcd(minus(x, y), y)
   if2(false, x, y) -> x
   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)
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) -> gcd(y, x)
   if2(true, x, y) -> gcd(minus(x, y), y)
   if2(false, x, y) -> x
   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)
   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)
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) -> GCD(y, x)
   IF2(true, x, y) -> GCD(minus(x, y), y)
   IF2(true, x, y) -> MINUS(x, y)
   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) -> gcd(y, x)
   if2(true, x, y) -> gcd(minus(x, y), y)
   if2(false, x, y) -> x
   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)
   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)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 4 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) -> gcd(y, x)
   if2(true, x, y) -> gcd(minus(x, y), y)
   if2(false, x, y) -> x
   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)
   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)
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)
   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)
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)
   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)
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) -> gcd(y, x)
   if2(true, x, y) -> gcd(minus(x, y), y)
   if2(false, x, y) -> x
   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)
   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)
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)
   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)
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)
   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)
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) -> gcd(y, x)
   if2(true, x, y) -> gcd(minus(x, y), y)
   if2(false, x, y) -> x
   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)
   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)
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)
   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)
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)
   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)
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) -> GCD(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) -> gcd(y, x)
   if2(true, x, y) -> gcd(minus(x, y), y)
   if2(false, x, y) -> x
   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)
   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)
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) -> GCD(y, x)

The TRS R consists of the following rules:

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

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)
   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)
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)


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

(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) -> GCD(y, x)

The TRS R consists of the following rules:

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

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) NonInfProof (EQUIVALENT)
The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:

Note that final constraints are written in bold face.



For Pair IF2(true, x, y) -> GCD(minus(x, y), y) the following chains were created:
*We consider the chain IF1(true, x4, x5) -> IF2(gt(x5, 0), x4, x5), IF2(true, x6, x7) -> GCD(minus(x6, x7), x7) which results in the following constraint:

(1)    (IF2(gt(x5, 0), x4, x5)=IF2(true, x6, x7)  ==>  IF2(true, x6, x7)_>=_GCD(minus(x6, x7), x7))



We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint:

(2)    (0=x42 & gt(x5, x42)=true  ==>  IF2(true, x4, x5)_>=_GCD(minus(x4, x5), x5))



We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gt(x5, x42)=true which results in the following new constraints:

(3)    (true=true & 0=0  ==>  IF2(true, x4, s(x44))_>=_GCD(minus(x4, s(x44)), s(x44)))

(4)    (gt(x46, x45)=true & 0=s(x45) & (\/x47:gt(x46, x45)=true & 0=x45  ==>  IF2(true, x47, x46)_>=_GCD(minus(x47, x46), x46))  ==>  IF2(true, x4, s(x46))_>=_GCD(minus(x4, s(x46)), s(x46)))



We simplified constraint (3) using rules (I), (II) which results in the following new constraint:

(5)    (IF2(true, x4, s(x44))_>=_GCD(minus(x4, s(x44)), s(x44)))



We solved constraint (4) using rules (I), (II).




For Pair GCD(x, y) -> IF1(ge(x, y), x, y) the following chains were created:
*We consider the chain IF2(true, x10, x11) -> GCD(minus(x10, x11), x11), GCD(x12, x13) -> IF1(ge(x12, x13), x12, x13) which results in the following constraint:

(1)    (GCD(minus(x10, x11), x11)=GCD(x12, x13)  ==>  GCD(x12, x13)_>=_IF1(ge(x12, x13), x12, x13))



We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint:

(2)    (GCD(x12, x11)_>=_IF1(ge(x12, x11), x12, x11))




*We consider the chain IF1(false, x18, x19) -> GCD(x19, x18), GCD(x20, x21) -> IF1(ge(x20, x21), x20, x21) which results in the following constraint:

(1)    (GCD(x19, x18)=GCD(x20, x21)  ==>  GCD(x20, x21)_>=_IF1(ge(x20, x21), x20, x21))



We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:

(2)    (GCD(x19, x18)_>=_IF1(ge(x19, x18), x19, x18))








For Pair IF1(true, x, y) -> IF2(gt(y, 0), x, y) the following chains were created:
*We consider the chain GCD(x24, x25) -> IF1(ge(x24, x25), x24, x25), IF1(true, x26, x27) -> IF2(gt(x27, 0), x26, x27) which results in the following constraint:

(1)    (IF1(ge(x24, x25), x24, x25)=IF1(true, x26, x27)  ==>  IF1(true, x26, x27)_>=_IF2(gt(x27, 0), x26, x27))



We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:

(2)    (ge(x24, x25)=true  ==>  IF1(true, x24, x25)_>=_IF2(gt(x25, 0), x24, x25))



We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on ge(x24, x25)=true which results in the following new constraints:

(3)    (true=true  ==>  IF1(true, x48, 0)_>=_IF2(gt(0, 0), x48, 0))

(4)    (ge(x51, x50)=true & (ge(x51, x50)=true  ==>  IF1(true, x51, x50)_>=_IF2(gt(x50, 0), x51, x50))  ==>  IF1(true, s(x51), s(x50))_>=_IF2(gt(s(x50), 0), s(x51), s(x50)))



We simplified constraint (3) using rules (I), (II) which results in the following new constraint:

(5)    (IF1(true, x48, 0)_>=_IF2(gt(0, 0), x48, 0))



We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (ge(x51, x50)=true  ==>  IF1(true, x51, x50)_>=_IF2(gt(x50, 0), x51, x50)) with sigma = [ ] which results in the following new constraint:

(6)    (IF1(true, x51, x50)_>=_IF2(gt(x50, 0), x51, x50)  ==>  IF1(true, s(x51), s(x50))_>=_IF2(gt(s(x50), 0), s(x51), s(x50)))








For Pair IF1(false, x, y) -> GCD(y, x) the following chains were created:
*We consider the chain GCD(x34, x35) -> IF1(ge(x34, x35), x34, x35), IF1(false, x36, x37) -> GCD(x37, x36) which results in the following constraint:

(1)    (IF1(ge(x34, x35), x34, x35)=IF1(false, x36, x37)  ==>  IF1(false, x36, x37)_>=_GCD(x37, x36))



We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:

(2)    (ge(x34, x35)=false  ==>  IF1(false, x34, x35)_>=_GCD(x35, x34))



We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on ge(x34, x35)=false which results in the following new constraints:

(3)    (false=false  ==>  IF1(false, 0, s(x53))_>=_GCD(s(x53), 0))

(4)    (ge(x55, x54)=false & (ge(x55, x54)=false  ==>  IF1(false, x55, x54)_>=_GCD(x54, x55))  ==>  IF1(false, s(x55), s(x54))_>=_GCD(s(x54), s(x55)))



We simplified constraint (3) using rules (I), (II) which results in the following new constraint:

(5)    (IF1(false, 0, s(x53))_>=_GCD(s(x53), 0))



We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (ge(x55, x54)=false  ==>  IF1(false, x55, x54)_>=_GCD(x54, x55)) with sigma = [ ] which results in the following new constraint:

(6)    (IF1(false, x55, x54)_>=_GCD(x54, x55)  ==>  IF1(false, s(x55), s(x54))_>=_GCD(s(x54), s(x55)))








To summarize, we get the following constraints P__>=_ for the following pairs.

*IF2(true, x, y) -> GCD(minus(x, y), y)

*(IF2(true, x4, s(x44))_>=_GCD(minus(x4, s(x44)), s(x44)))




*GCD(x, y) -> IF1(ge(x, y), x, y)

*(GCD(x12, x11)_>=_IF1(ge(x12, x11), x12, x11))


*(GCD(x19, x18)_>=_IF1(ge(x19, x18), x19, x18))




*IF1(true, x, y) -> IF2(gt(y, 0), x, y)

*(IF1(true, x48, 0)_>=_IF2(gt(0, 0), x48, 0))


*(IF1(true, x51, x50)_>=_IF2(gt(x50, 0), x51, x50)  ==>  IF1(true, s(x51), s(x50))_>=_IF2(gt(s(x50), 0), s(x51), s(x50)))




*IF1(false, x, y) -> GCD(y, x)

*(IF1(false, 0, s(x53))_>=_GCD(s(x53), 0))


*(IF1(false, x55, x54)_>=_GCD(x54, x55)  ==>  IF1(false, s(x55), s(x54))_>=_GCD(s(x54), s(x55)))








The constraints for P_> respective P_bound are constructed from P__>=_ where we just replace every occurence of "t _>=_ s" in P__>=_ by  "t > s" respective "t _>=_ c". Here c stands for the fresh constant used for P_bound. 

Using the following integer polynomial ordering the  resulting constraints can be solved 

Polynomial interpretation [NONINF]:

   POL(0) = 0
   POL(GCD(x_1, x_2)) = -1 + x_2
   POL(IF1(x_1, x_2, x_3)) = -1 - x_1 + x_3
   POL(IF2(x_1, x_2, x_3)) = -1 - x_1 + x_3
   POL(c) = -1
   POL(false) = 0
   POL(ge(x_1, x_2)) = 0
   POL(gt(x_1, x_2)) = 0
   POL(if(x_1, x_2, x_3)) = 0
   POL(minus(x_1, x_2)) = 0
   POL(s(x_1)) = 1 + x_1
   POL(true) = 0


The following pairs  are in P_>:
   IF1(false, x, y) -> GCD(y, x)
The following pairs are in P_bound:
   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) -> GCD(y, x)
The following rules are usable:
   true -> ge(x, 0)
   false -> ge(0, s(x))
   ge(x, y) -> ge(s(x), s(y))
   false -> gt(0, y)
   true -> gt(s(x), 0)

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

(34)
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)

The TRS R consists of the following rules:

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

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)
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)))


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

(36)
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)
   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:

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

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

(38)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   GCD(x0, 0) -> IF1(true, x0, 0)
   IF1(true, x, y) -> IF2(gt(y, 0), x, y)
   IF2(true, x, y) -> GCD(minus(x, y), y)
   GCD(s(x0), s(x1)) -> IF1(ge(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

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

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)
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, 0) -> IF2(false, y0, 0),IF1(true, y0, 0) -> IF2(false, y0, 0))
   (IF1(true, y0, s(x0)) -> IF2(true, y0, s(x0)),IF1(true, y0, s(x0)) -> IF2(true, y0, s(x0)))


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

(40)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   GCD(x0, 0) -> IF1(true, x0, 0)
   IF2(true, x, y) -> GCD(minus(x, y), y)
   GCD(s(x0), s(x1)) -> IF1(ge(x0, x1), s(x0), s(x1))
   IF1(true, y0, 0) -> IF2(false, y0, 0)
   IF1(true, y0, s(x0)) -> IF2(true, y0, s(x0))

The TRS R consists of the following rules:

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

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)
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)

The TRS R consists of the following rules:

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

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)
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))
   IF2(true, s(x0), x1) -> GCD(if(gt(s(x0), x1), x0, x1), x1)

The TRS R consists of the following rules:

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

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)
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)))


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

(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))
   IF2(true, s(x0), x1) -> GCD(if(gt(s(x0), x1), x0, x1), x1)
   IF1(true, s(z0), s(z1)) -> IF2(true, s(z0), s(z1))

The TRS R consists of the following rules:

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

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)
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)))


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

(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(true, s(z0), s(z1)) -> IF2(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:

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

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)
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)))


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

(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))
   IF1(true, s(z0), s(z1)) -> IF2(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:

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

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) Induction-Processor (SOUND)

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(x_1, x_2)) = x_1
   POL(IF1(x_1, x_2, x_3)) = x_2
   POL(IF2(x_1, x_2, x_3)) = x_1 + x_2
   POL(false_renamed) = 0
   POL(ge(x_1, x_2)) = x_1
   POL(gt(x_1, x_2)) = 0
   POL(if(x_1, x_2, x_3)) = 1 + x_2
   POL(minus(x_1, x_2)) = x_1
   POL(s(x_1)) = 1 + x_1
   POL(true_renamed) = 0

At least one of these decreasing rules is always used after the deleted DP:
if(false_renamed, x5, y2) -> 0


The following formula is valid:
z0':sort[a23],z1':sort[a23].if'(gt(z0', z1'), z0', s(z1'))=true


The transformed set:
if'(true_renamed, x4, y1) -> minus'(x4, y1)
if'(false_renamed, x5, y2) -> true
minus'(s(x6), y3) -> if'(gt(s(x6), y3), x6, y3)
minus'(0, v43) -> false
gt(0, y) -> false_renamed
gt(s(x), 0) -> true_renamed
ge(x', 0) -> true_renamed
ge(0, s(x'')) -> false_renamed
ge(s(x2), s(y')) -> ge(x2, y')
gt(s(x3), s(y'')) -> gt(x3, y'')
if(true_renamed, x4, y1) -> s(minus(x4, y1))
if(false_renamed, x5, y2) -> 0
minus(s(x6), y3) -> if(gt(s(x6), y3), x6, y3)
minus(0, v43) -> 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[a23](0, 0) -> true
equal_sort[a23](0, s(v45)) -> false
equal_sort[a23](s(v46), 0) -> false
equal_sort[a23](s(v46), s(v47)) -> equal_sort[a23](v46, v47)
equal_sort[a22](false_renamed, false_renamed) -> true
equal_sort[a22](false_renamed, true_renamed) -> false
equal_sort[a22](true_renamed, false_renamed) -> false
equal_sort[a22](true_renamed, true_renamed) -> true
equal_sort[a40](witness_sort[a40], witness_sort[a40]) -> true


The proof given by the theorem prover:
The following output was given by the internal theorem prover:proof of internal
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Partial correctness of the following Program

   [x, v45, v46, v47, x5, y2, x6, y1, v43, y3, y, x3, y'', x', x'', x2, y']
   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[a23](0, 0) -> true
   equal_sort[a23](0, s(v45)) -> false
   equal_sort[a23](s(v46), 0) -> false
   equal_sort[a23](s(v46), s(v47)) -> equal_sort[a23](v46, v47)
   equal_sort[a22](false_renamed, false_renamed) -> true
   equal_sort[a22](false_renamed, true_renamed) -> false
   equal_sort[a22](true_renamed, false_renamed) -> false
   equal_sort[a22](true_renamed, true_renamed) -> true
   equal_sort[a40](witness_sort[a40], witness_sort[a40]) -> true
   if'(false_renamed, x5, y2) -> true
   if'(true_renamed, s(x6), y1) -> if'(gt(s(x6), y1), x6, y1)
   if'(true_renamed, 0, y1) -> false
   minus'(0, v43) -> 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(0, y) -> false_renamed
   gt(s(x), 0) -> true_renamed
   gt(s(x3), s(y'')) -> gt(x3, y'')
   ge(x', 0) -> true_renamed
   ge(0, s(x'')) -> false_renamed
   ge(s(x2), s(y')) -> ge(x2, y')
   if(false_renamed, x5, y2) -> 0
   if(true_renamed, s(x6), y1) -> s(if(gt(s(x6), y1), x6, y1))
   if(true_renamed, 0, y1) -> s(0)
   minus(0, v43) -> 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:
z0':sort[a23],z1':sort[a23].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) YES
    (6) Formula
        (7) Symbolic evaluation [EQUIVALENT, 0 ms]
        (8) Formula
        (9) Induction by data structure [EQUIVALENT, 0 ms]
        (10) AND
            (11) Formula
                (12) Symbolic evaluation [EQUIVALENT, 0 ms]
                (13) YES
            (14) Formula
                (15) Symbolic evaluation under hypothesis [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[a23],z1':sort[a23].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:
y:sort[a23].if'(gt(0, y), 0, s(y))=true

There are no hypotheses.





2. Base Case:
Formula:
x:sort[a23].if'(gt(s(x), 0), s(x), s(0))=true

There are no hypotheses.





1. Step Case:
Formula:
x3:sort[a23],y'':sort[a23].if'(gt(s(x3), s(y'')), s(x3), s(s(y'')))=true

Hypotheses:
x3:sort[a23],y'':sort[a23].if'(gt(x3, y''), x3, s(y''))=true






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

(2)
Complex Obligation (AND)

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

(3)
Obligation:
Formula:
y:sort[a23].if'(gt(0, y), 0, s(y))=true

There are no hypotheses.




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

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

(5)
YES

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

(6)
Obligation:
Formula:
x:sort[a23].if'(gt(s(x), 0), s(x), s(0))=true

There are no hypotheses.




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

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

(8)
Obligation:
Formula:
x:sort[a23].if'(gt(x, 0), x, s(0))=true

There are no hypotheses.




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

(9) Induction by data structure (EQUIVALENT)
Induction by data structure sort[a23] 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[a23].if'(gt(s(n), 0), s(n), s(0))=true

Hypotheses:
n:sort[a23].if'(gt(n, 0), n, s(0))=true






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

(10)
Complex Obligation (AND)

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

(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:
n:sort[a23].if'(gt(s(n), 0), s(n), s(0))=true

Hypotheses:
n:sort[a23].if'(gt(n, 0), n, s(0))=true




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

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

n:sort[a23].if'(gt(n, 0), n, s(0))=true

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

(16)
YES

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

(17)
Obligation:
Formula:
x3:sort[a23],y'':sort[a23].if'(gt(s(x3), s(y'')), s(x3), s(s(y'')))=true

Hypotheses:
x3:sort[a23],y'':sort[a23].if'(gt(x3, y''), x3, s(y''))=true




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

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

(19)
Obligation:
Formula:
x3:sort[a23],y'':sort[a23].if'(gt(x3, y''), s(x3), s(s(y'')))=true

Hypotheses:
x3:sort[a23],y'':sort[a23].if'(gt(x3, y''), x3, s(y''))=true




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

(20) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
x3:sort[a23],y'':sort[a23].(if'(gt(x3, y''), x3, s(y''))=true->if'(gt(x3, y''), s(x3), s(s(y'')))=true)

There are no hypotheses.




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

(21)
Obligation:
Formula:
x3:sort[a23],y'':sort[a23].(if'(gt(x3, y''), x3, s(y''))=true->if'(gt(x3, y''), s(x3), s(s(y'')))=true)

There are no hypotheses.




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

(22) Inverse Substitution (SOUND)
The formula could be generalised by inverse substitution to:
n:sort[a22],x3:sort[a23],y'':sort[a23].(if'(n, x3, s(y''))=true->if'(n, s(x3), s(s(y'')))=true)

Inverse substitution used:
[gt(x3, y'')/n]


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

(23)
Obligation:
Formula:
n:sort[a22],x3:sort[a23],y'':sort[a23].(if'(n, x3, s(y''))=true->if'(n, s(x3), s(s(y'')))=true)

There are no hypotheses.




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

(24) Inverse Substitution (SOUND)
The formula could be generalised by inverse substitution to:
n:sort[a22],x3:sort[a23],n':sort[a23].(if'(n, x3, n')=true->if'(n, s(x3), s(n'))=true)

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


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

(25)
Obligation:
Formula:
n:sort[a22],x3:sort[a23],n':sort[a23].(if'(n, x3, n')=true->if'(n, s(x3), s(n'))=true)

There are no hypotheses.




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

(26) Induction by algorithm (EQUIVALENT)
Induction by algorithm if'(n, x3, n') generates the following cases:

1. Base Case:
Formula:
x5:sort[a23],y2:sort[a23].(if'(false_renamed, x5, y2)=true->if'(false_renamed, s(x5), s(y2))=true)

There are no hypotheses.





2. Base Case:
Formula:
y1:sort[a23].(if'(true_renamed, 0, y1)=true->if'(true_renamed, s(0), s(y1))=true)

There are no hypotheses.





1. Step Case:
Formula:
x6:sort[a23],y1:sort[a23].(if'(true_renamed, s(x6), y1)=true->if'(true_renamed, s(s(x6)), s(y1))=true)

Hypotheses:
x6:sort[a23],y1:sort[a23].(if'(gt(s(x6), y1), x6, y1)=true->if'(gt(s(x6), y1), s(x6), s(y1))=true)






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

(27)
Complex Obligation (AND)

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

(28)
Obligation:
Formula:
x5:sort[a23],y2:sort[a23].(if'(false_renamed, x5, y2)=true->if'(false_renamed, s(x5), s(y2))=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:
y1:sort[a23].(if'(true_renamed, 0, y1)=true->if'(true_renamed, s(0), s(y1))=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[a23],y1:sort[a23].(if'(true_renamed, s(x6), y1)=true->if'(true_renamed, s(s(x6)), s(y1))=true)

Hypotheses:
x6:sort[a23],y1:sort[a23].(if'(gt(s(x6), y1), x6, y1)=true->if'(gt(s(x6), y1), s(x6), s(y1))=true)




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

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

x6:sort[a23],y1:sort[a23].(if'(gt(s(x6), y1), x6, y1)=true->if'(gt(s(x6), y1), s(x6), s(y1))=true)

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

(36)
YES

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

(52)
Complex Obligation (AND)

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

(53)
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))

The TRS R consists of the following rules:

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

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.
----------------------------------------

(54) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
----------------------------------------

(55)
TRUE

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

(56)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   if'(true_renamed, x4, y1) -> minus'(x4, y1)
   if'(false_renamed, x5, y2) -> true
   minus'(s(x6), y3) -> if'(gt(s(x6), y3), x6, y3)
   minus'(0, v43) -> false
   gt(0, y) -> false_renamed
   gt(s(x), 0) -> true_renamed
   ge(x', 0) -> true_renamed
   ge(0, s(x'')) -> false_renamed
   ge(s(x2), s(y')) -> ge(x2, y')
   gt(s(x3), s(y'')) -> gt(x3, y'')
   if(true_renamed, x4, y1) -> s(minus(x4, y1))
   if(false_renamed, x5, y2) -> 0
   minus(s(x6), y3) -> if(gt(s(x6), y3), x6, y3)
   minus(0, v43) -> 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[a23](0, 0) -> true
   equal_sort[a23](0, s(v45)) -> false
   equal_sort[a23](s(v46), 0) -> false
   equal_sort[a23](s(v46), s(v47)) -> equal_sort[a23](v46, v47)
   equal_sort[a22](false_renamed, false_renamed) -> true
   equal_sort[a22](false_renamed, true_renamed) -> false
   equal_sort[a22](true_renamed, false_renamed) -> false
   equal_sort[a22](true_renamed, true_renamed) -> true
   equal_sort[a40](witness_sort[a40], witness_sort[a40]) -> true

Q is empty.

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

(57) QTRSRRRProof (EQUIVALENT)
Used ordering:
Quasi precedence:
ge_2 > true_renamed
ge_2 > false_renamed
[if_3, minus_2] > [s_1, false] > true_renamed
[if_3, minus_2] > [s_1, false] > false_renamed
[if_3, minus_2] > [s_1, false] > gt_2
[if_3, minus_2] > 0 > false_renamed
and_2 > [s_1, false] > true_renamed
and_2 > [s_1, false] > false_renamed
and_2 > [s_1, false] > gt_2
not_1 > [if'_3, minus'_2, true, equal_bool_2] > [s_1, false] > true_renamed
not_1 > [if'_3, minus'_2, true, equal_bool_2] > [s_1, false] > false_renamed
not_1 > [if'_3, minus'_2, true, equal_bool_2] > [s_1, false] > gt_2
isa_false_1 > [if'_3, minus'_2, true, equal_bool_2] > [s_1, false] > true_renamed
isa_false_1 > [if'_3, minus'_2, true, equal_bool_2] > [s_1, false] > false_renamed
isa_false_1 > [if'_3, minus'_2, true, equal_bool_2] > [s_1, false] > gt_2
equal_sort[a23]_2 > [if'_3, minus'_2, true, equal_bool_2] > [s_1, false] > true_renamed
equal_sort[a23]_2 > [if'_3, minus'_2, true, equal_bool_2] > [s_1, false] > false_renamed
equal_sort[a23]_2 > [if'_3, minus'_2, true, equal_bool_2] > [s_1, false] > gt_2
equal_sort[a22]_2 > [if'_3, minus'_2, true, equal_bool_2] > [s_1, false] > true_renamed
equal_sort[a22]_2 > [if'_3, minus'_2, true, equal_bool_2] > [s_1, false] > false_renamed
equal_sort[a22]_2 > [if'_3, minus'_2, true, equal_bool_2] > [s_1, false] > gt_2
witness_sort[a40] > [if'_3, minus'_2, true, equal_bool_2] > [s_1, false] > true_renamed
witness_sort[a40] > [if'_3, minus'_2, true, equal_bool_2] > [s_1, false] > false_renamed
witness_sort[a40] > [if'_3, minus'_2, true, equal_bool_2] > [s_1, false] > gt_2


Status:
if'_3: [3,2,1]
true_renamed: multiset status
minus'_2: [2,1]
false_renamed: multiset status
true: multiset status
s_1: multiset status
gt_2: multiset status
0: multiset status
false: multiset status
ge_2: [2,1]
if_3: [3,2,1]
minus_2: [2,1]
equal_bool_2: [2,1]
and_2: [2,1]
or_2: [2,1]
not_1: [1]
isa_true_1: multiset status
isa_false_1: [1]
equal_sort[a23]_2: [2,1]
equal_sort[a22]_2: [2,1]
equal_sort[a40]_2: [2,1]
witness_sort[a40]: multiset status

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

   if'(true_renamed, x4, y1) -> minus'(x4, y1)
   if'(false_renamed, x5, y2) -> true
   minus'(s(x6), y3) -> if'(gt(s(x6), y3), x6, y3)
   minus'(0, v43) -> false
   gt(0, y) -> false_renamed
   gt(s(x), 0) -> true_renamed
   ge(x', 0) -> true_renamed
   ge(0, s(x'')) -> false_renamed
   ge(s(x2), s(y')) -> ge(x2, y')
   gt(s(x3), s(y'')) -> gt(x3, y'')
   if(true_renamed, x4, y1) -> s(minus(x4, y1))
   if(false_renamed, x5, y2) -> 0
   minus(s(x6), y3) -> if(gt(s(x6), y3), x6, y3)
   minus(0, v43) -> 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[a23](0, 0) -> true
   equal_sort[a23](0, s(v45)) -> false
   equal_sort[a23](s(v46), 0) -> false
   equal_sort[a23](s(v46), s(v47)) -> equal_sort[a23](v46, v47)
   equal_sort[a22](false_renamed, false_renamed) -> true
   equal_sort[a22](false_renamed, true_renamed) -> false
   equal_sort[a22](true_renamed, false_renamed) -> false
   equal_sort[a22](true_renamed, true_renamed) -> true
   equal_sort[a40](witness_sort[a40], witness_sort[a40]) -> true




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(58)
Obligation:
Q restricted rewrite system:
R is empty.
Q is empty.

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(59) RisEmptyProof (EQUIVALENT)
The TRS R is empty. Hence, termination is trivially proven.
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(60)
YES