/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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


YES
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 1 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) TransformationProof [EQUIVALENT, 0 ms]
        (27) QDP
        (28) UsableRulesProof [EQUIVALENT, 0 ms]
        (29) QDP
        (30) QReductionProof [EQUIVALENT, 0 ms]
        (31) QDP
        (32) NonInfProof [EQUIVALENT, 61 ms]
        (33) AND
            (34) QDP
                (35) DependencyGraphProof [EQUIVALENT, 0 ms]
                (36) TRUE
            (37) QDP
                (38) DependencyGraphProof [EQUIVALENT, 0 ms]
                (39) TRUE


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

(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   gt(s(x), 0) -> true
   gt(0, y) -> false
   gt(s(x), s(y)) -> gt(x, y)
   divides(x, y) -> div(x, y, y)
   div(0, 0, z) -> true
   div(0, s(x), z) -> false
   div(s(x), 0, s(z)) -> div(s(x), s(z), s(z))
   div(s(x), s(y), z) -> div(x, y, z)
   prime(x) -> test(x, s(s(0)))
   test(x, y) -> if1(gt(x, y), x, y)
   if1(true, x, y) -> if2(divides(x, y), x, y)
   if1(false, x, y) -> true
   if2(true, x, y) -> false
   if2(false, x, y) -> test(x, s(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:

   gt(s(x), 0) -> true
   gt(0, y) -> false
   gt(s(x), s(y)) -> gt(x, y)
   divides(x, y) -> div(x, y, y)
   div(0, 0, z) -> true
   div(0, s(x), z) -> false
   div(s(x), 0, s(z)) -> div(s(x), s(z), s(z))
   div(s(x), s(y), z) -> div(x, y, z)
   prime(x) -> test(x, s(s(0)))
   test(x, y) -> if1(gt(x, y), x, y)
   if1(true, x, y) -> if2(divides(x, y), x, y)
   if1(false, x, y) -> true
   if2(true, x, y) -> false
   if2(false, x, y) -> test(x, s(y))

The set Q consists of the following terms:

   gt(s(x0), 0)
   gt(0, x0)
   gt(s(x0), s(x1))
   divides(x0, x1)
   div(0, 0, x0)
   div(0, s(x0), x1)
   div(s(x0), 0, s(x1))
   div(s(x0), s(x1), x2)
   prime(x0)
   test(x0, x1)
   if1(true, x0, x1)
   if1(false, x0, x1)
   if2(true, x0, x1)
   if2(false, x0, 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:

   GT(s(x), s(y)) -> GT(x, y)
   DIVIDES(x, y) -> DIV(x, y, y)
   DIV(s(x), 0, s(z)) -> DIV(s(x), s(z), s(z))
   DIV(s(x), s(y), z) -> DIV(x, y, z)
   PRIME(x) -> TEST(x, s(s(0)))
   TEST(x, y) -> IF1(gt(x, y), x, y)
   TEST(x, y) -> GT(x, y)
   IF1(true, x, y) -> IF2(divides(x, y), x, y)
   IF1(true, x, y) -> DIVIDES(x, y)
   IF2(false, x, y) -> TEST(x, s(y))

The TRS R consists of the following rules:

   gt(s(x), 0) -> true
   gt(0, y) -> false
   gt(s(x), s(y)) -> gt(x, y)
   divides(x, y) -> div(x, y, y)
   div(0, 0, z) -> true
   div(0, s(x), z) -> false
   div(s(x), 0, s(z)) -> div(s(x), s(z), s(z))
   div(s(x), s(y), z) -> div(x, y, z)
   prime(x) -> test(x, s(s(0)))
   test(x, y) -> if1(gt(x, y), x, y)
   if1(true, x, y) -> if2(divides(x, y), x, y)
   if1(false, x, y) -> true
   if2(true, x, y) -> false
   if2(false, x, y) -> test(x, s(y))

The set Q consists of the following terms:

   gt(s(x0), 0)
   gt(0, x0)
   gt(s(x0), s(x1))
   divides(x0, x1)
   div(0, 0, x0)
   div(0, s(x0), x1)
   div(s(x0), 0, s(x1))
   div(s(x0), s(x1), x2)
   prime(x0)
   test(x0, x1)
   if1(true, x0, x1)
   if1(false, x0, x1)
   if2(true, x0, x1)
   if2(false, x0, x1)

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 4 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

(7)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   DIV(s(x), s(y), z) -> DIV(x, y, z)
   DIV(s(x), 0, s(z)) -> DIV(s(x), s(z), s(z))

The TRS R consists of the following rules:

   gt(s(x), 0) -> true
   gt(0, y) -> false
   gt(s(x), s(y)) -> gt(x, y)
   divides(x, y) -> div(x, y, y)
   div(0, 0, z) -> true
   div(0, s(x), z) -> false
   div(s(x), 0, s(z)) -> div(s(x), s(z), s(z))
   div(s(x), s(y), z) -> div(x, y, z)
   prime(x) -> test(x, s(s(0)))
   test(x, y) -> if1(gt(x, y), x, y)
   if1(true, x, y) -> if2(divides(x, y), x, y)
   if1(false, x, y) -> true
   if2(true, x, y) -> false
   if2(false, x, y) -> test(x, s(y))

The set Q consists of the following terms:

   gt(s(x0), 0)
   gt(0, x0)
   gt(s(x0), s(x1))
   divides(x0, x1)
   div(0, 0, x0)
   div(0, s(x0), x1)
   div(s(x0), 0, s(x1))
   div(s(x0), s(x1), x2)
   prime(x0)
   test(x0, x1)
   if1(true, x0, x1)
   if1(false, x0, x1)
   if2(true, x0, x1)
   if2(false, x0, 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:

   DIV(s(x), s(y), z) -> DIV(x, y, z)
   DIV(s(x), 0, s(z)) -> DIV(s(x), s(z), s(z))

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

   gt(s(x0), 0)
   gt(0, x0)
   gt(s(x0), s(x1))
   divides(x0, x1)
   div(0, 0, x0)
   div(0, s(x0), x1)
   div(s(x0), 0, s(x1))
   div(s(x0), s(x1), x2)
   prime(x0)
   test(x0, x1)
   if1(true, x0, x1)
   if1(false, x0, x1)
   if2(true, x0, x1)
   if2(false, x0, 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].

   gt(s(x0), 0)
   gt(0, x0)
   gt(s(x0), s(x1))
   divides(x0, x1)
   div(0, 0, x0)
   div(0, s(x0), x1)
   div(s(x0), 0, s(x1))
   div(s(x0), s(x1), x2)
   prime(x0)
   test(x0, x1)
   if1(true, x0, x1)
   if1(false, x0, x1)
   if2(true, x0, x1)
   if2(false, x0, x1)


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

(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   DIV(s(x), s(y), z) -> DIV(x, y, z)
   DIV(s(x), 0, s(z)) -> DIV(s(x), s(z), s(z))

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:
*DIV(s(x), s(y), z) -> DIV(x, y, z)
The graph contains the following edges 1 > 1, 2 > 2, 3 >= 3


*DIV(s(x), 0, s(z)) -> DIV(s(x), s(z), s(z))
The graph contains the following edges 1 >= 1, 3 >= 2, 3 >= 3


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

(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:

   gt(s(x), 0) -> true
   gt(0, y) -> false
   gt(s(x), s(y)) -> gt(x, y)
   divides(x, y) -> div(x, y, y)
   div(0, 0, z) -> true
   div(0, s(x), z) -> false
   div(s(x), 0, s(z)) -> div(s(x), s(z), s(z))
   div(s(x), s(y), z) -> div(x, y, z)
   prime(x) -> test(x, s(s(0)))
   test(x, y) -> if1(gt(x, y), x, y)
   if1(true, x, y) -> if2(divides(x, y), x, y)
   if1(false, x, y) -> true
   if2(true, x, y) -> false
   if2(false, x, y) -> test(x, s(y))

The set Q consists of the following terms:

   gt(s(x0), 0)
   gt(0, x0)
   gt(s(x0), s(x1))
   divides(x0, x1)
   div(0, 0, x0)
   div(0, s(x0), x1)
   div(s(x0), 0, s(x1))
   div(s(x0), s(x1), x2)
   prime(x0)
   test(x0, x1)
   if1(true, x0, x1)
   if1(false, x0, x1)
   if2(true, x0, x1)
   if2(false, x0, 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:

   gt(s(x0), 0)
   gt(0, x0)
   gt(s(x0), s(x1))
   divides(x0, x1)
   div(0, 0, x0)
   div(0, s(x0), x1)
   div(s(x0), 0, s(x1))
   div(s(x0), s(x1), x2)
   prime(x0)
   test(x0, x1)
   if1(true, x0, x1)
   if1(false, x0, x1)
   if2(true, x0, x1)
   if2(false, x0, 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].

   gt(s(x0), 0)
   gt(0, x0)
   gt(s(x0), s(x1))
   divides(x0, x1)
   div(0, 0, x0)
   div(0, s(x0), x1)
   div(s(x0), 0, s(x1))
   div(s(x0), s(x1), x2)
   prime(x0)
   test(x0, x1)
   if1(true, x0, x1)
   if1(false, x0, x1)
   if2(true, x0, x1)
   if2(false, x0, 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:

   TEST(x, y) -> IF1(gt(x, y), x, y)
   IF1(true, x, y) -> IF2(divides(x, y), x, y)
   IF2(false, x, y) -> TEST(x, s(y))

The TRS R consists of the following rules:

   gt(s(x), 0) -> true
   gt(0, y) -> false
   gt(s(x), s(y)) -> gt(x, y)
   divides(x, y) -> div(x, y, y)
   div(0, 0, z) -> true
   div(0, s(x), z) -> false
   div(s(x), 0, s(z)) -> div(s(x), s(z), s(z))
   div(s(x), s(y), z) -> div(x, y, z)
   prime(x) -> test(x, s(s(0)))
   test(x, y) -> if1(gt(x, y), x, y)
   if1(true, x, y) -> if2(divides(x, y), x, y)
   if1(false, x, y) -> true
   if2(true, x, y) -> false
   if2(false, x, y) -> test(x, s(y))

The set Q consists of the following terms:

   gt(s(x0), 0)
   gt(0, x0)
   gt(s(x0), s(x1))
   divides(x0, x1)
   div(0, 0, x0)
   div(0, s(x0), x1)
   div(s(x0), 0, s(x1))
   div(s(x0), s(x1), x2)
   prime(x0)
   test(x0, x1)
   if1(true, x0, x1)
   if1(false, x0, x1)
   if2(true, x0, x1)
   if2(false, x0, 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:

   TEST(x, y) -> IF1(gt(x, y), x, y)
   IF1(true, x, y) -> IF2(divides(x, y), x, y)
   IF2(false, x, y) -> TEST(x, s(y))

The TRS R consists of the following rules:

   divides(x, y) -> div(x, y, y)
   div(0, 0, z) -> true
   div(0, s(x), z) -> false
   div(s(x), 0, s(z)) -> div(s(x), s(z), s(z))
   div(s(x), s(y), z) -> div(x, y, z)
   gt(s(x), 0) -> true
   gt(0, y) -> false
   gt(s(x), s(y)) -> gt(x, y)

The set Q consists of the following terms:

   gt(s(x0), 0)
   gt(0, x0)
   gt(s(x0), s(x1))
   divides(x0, x1)
   div(0, 0, x0)
   div(0, s(x0), x1)
   div(s(x0), 0, s(x1))
   div(s(x0), s(x1), x2)
   prime(x0)
   test(x0, x1)
   if1(true, x0, x1)
   if1(false, x0, x1)
   if2(true, x0, x1)
   if2(false, x0, 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].

   prime(x0)
   test(x0, x1)
   if1(true, x0, x1)
   if1(false, x0, x1)
   if2(true, x0, x1)
   if2(false, x0, x1)


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

(25)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TEST(x, y) -> IF1(gt(x, y), x, y)
   IF1(true, x, y) -> IF2(divides(x, y), x, y)
   IF2(false, x, y) -> TEST(x, s(y))

The TRS R consists of the following rules:

   divides(x, y) -> div(x, y, y)
   div(0, 0, z) -> true
   div(0, s(x), z) -> false
   div(s(x), 0, s(z)) -> div(s(x), s(z), s(z))
   div(s(x), s(y), z) -> div(x, y, z)
   gt(s(x), 0) -> true
   gt(0, y) -> false
   gt(s(x), s(y)) -> gt(x, y)

The set Q consists of the following terms:

   gt(s(x0), 0)
   gt(0, x0)
   gt(s(x0), s(x1))
   divides(x0, x1)
   div(0, 0, x0)
   div(0, s(x0), x1)
   div(s(x0), 0, s(x1))
   div(s(x0), s(x1), x2)

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

(26) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule IF1(true, x, y) -> IF2(divides(x, y), x, y) at position [0] we obtained the following new rules [LPAR04]:

   (IF1(true, x, y) -> IF2(div(x, y, y), x, y),IF1(true, x, y) -> IF2(div(x, y, y), x, y))


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

(27)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TEST(x, y) -> IF1(gt(x, y), x, y)
   IF2(false, x, y) -> TEST(x, s(y))
   IF1(true, x, y) -> IF2(div(x, y, y), x, y)

The TRS R consists of the following rules:

   divides(x, y) -> div(x, y, y)
   div(0, 0, z) -> true
   div(0, s(x), z) -> false
   div(s(x), 0, s(z)) -> div(s(x), s(z), s(z))
   div(s(x), s(y), z) -> div(x, y, z)
   gt(s(x), 0) -> true
   gt(0, y) -> false
   gt(s(x), s(y)) -> gt(x, y)

The set Q consists of the following terms:

   gt(s(x0), 0)
   gt(0, x0)
   gt(s(x0), s(x1))
   divides(x0, x1)
   div(0, 0, x0)
   div(0, s(x0), x1)
   div(s(x0), 0, s(x1))
   div(s(x0), s(x1), x2)

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

(28) 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.
----------------------------------------

(29)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TEST(x, y) -> IF1(gt(x, y), x, y)
   IF2(false, x, y) -> TEST(x, s(y))
   IF1(true, x, y) -> IF2(div(x, y, y), x, y)

The TRS R consists of the following rules:

   div(0, 0, z) -> true
   div(0, s(x), z) -> false
   div(s(x), 0, s(z)) -> div(s(x), s(z), s(z))
   div(s(x), s(y), z) -> div(x, y, z)
   gt(s(x), 0) -> true
   gt(0, y) -> false
   gt(s(x), s(y)) -> gt(x, y)

The set Q consists of the following terms:

   gt(s(x0), 0)
   gt(0, x0)
   gt(s(x0), s(x1))
   divides(x0, x1)
   div(0, 0, x0)
   div(0, s(x0), x1)
   div(s(x0), 0, s(x1))
   div(s(x0), s(x1), x2)

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

(30) 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].

   divides(x0, x1)


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

(31)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TEST(x, y) -> IF1(gt(x, y), x, y)
   IF2(false, x, y) -> TEST(x, s(y))
   IF1(true, x, y) -> IF2(div(x, y, y), x, y)

The TRS R consists of the following rules:

   div(0, 0, z) -> true
   div(0, s(x), z) -> false
   div(s(x), 0, s(z)) -> div(s(x), s(z), s(z))
   div(s(x), s(y), z) -> div(x, y, z)
   gt(s(x), 0) -> true
   gt(0, y) -> false
   gt(s(x), s(y)) -> gt(x, y)

The set Q consists of the following terms:

   gt(s(x0), 0)
   gt(0, x0)
   gt(s(x0), s(x1))
   div(0, 0, x0)
   div(0, s(x0), x1)
   div(s(x0), 0, s(x1))
   div(s(x0), s(x1), x2)

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

(32) 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 TEST(x, y) -> IF1(gt(x, y), x, y) the following chains were created:
*We consider the chain IF2(false, x2, x3) -> TEST(x2, s(x3)), TEST(x4, x5) -> IF1(gt(x4, x5), x4, x5) which results in the following constraint:

(1)    (TEST(x2, s(x3))=TEST(x4, x5)  ==>  TEST(x4, x5)_>=_IF1(gt(x4, x5), x4, x5))



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

(2)    (TEST(x2, s(x3))_>=_IF1(gt(x2, s(x3)), x2, s(x3)))








For Pair IF2(false, x, y) -> TEST(x, s(y)) the following chains were created:
*We consider the chain IF1(true, x12, x13) -> IF2(div(x12, x13, x13), x12, x13), IF2(false, x14, x15) -> TEST(x14, s(x15)) which results in the following constraint:

(1)    (IF2(div(x12, x13, x13), x12, x13)=IF2(false, x14, x15)  ==>  IF2(false, x14, x15)_>=_TEST(x14, s(x15)))



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

(2)    (x13=x24 & div(x12, x13, x24)=false  ==>  IF2(false, x12, x13)_>=_TEST(x12, s(x13)))



We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on div(x12, x13, x24)=false which results in the following new constraints:

(3)    (false=false & s(x27)=x26  ==>  IF2(false, 0, s(x27))_>=_TEST(0, s(s(x27))))

(4)    (div(s(x29), s(x28), s(x28))=false & 0=s(x28) & (div(s(x29), s(x28), s(x28))=false & s(x28)=s(x28)  ==>  IF2(false, s(x29), s(x28))_>=_TEST(s(x29), s(s(x28))))  ==>  IF2(false, s(x29), 0)_>=_TEST(s(x29), s(0)))

(5)    (div(x32, x31, x30)=false & s(x31)=x30 & (div(x32, x31, x30)=false & x31=x30  ==>  IF2(false, x32, x31)_>=_TEST(x32, s(x31)))  ==>  IF2(false, s(x32), s(x31))_>=_TEST(s(x32), s(s(x31))))



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

(6)    (IF2(false, 0, s(x27))_>=_TEST(0, s(s(x27))))



We solved constraint (4) using rules (I), (II).We simplified constraint (5) using rule (IV) which results in the following new constraint:

(7)    (div(x32, x31, x30)=false & s(x31)=x30  ==>  IF2(false, s(x32), s(x31))_>=_TEST(s(x32), s(s(x31))))



We simplified constraint (7) using rule (V) (with possible (I) afterwards) using induction on div(x32, x31, x30)=false which results in the following new constraints:

(8)    (false=false & s(s(x35))=x34  ==>  IF2(false, s(0), s(s(x35)))_>=_TEST(s(0), s(s(s(x35)))))

(9)    (div(s(x37), s(x36), s(x36))=false & s(0)=s(x36) & (div(s(x37), s(x36), s(x36))=false & s(s(x36))=s(x36)  ==>  IF2(false, s(s(x37)), s(s(x36)))_>=_TEST(s(s(x37)), s(s(s(x36)))))  ==>  IF2(false, s(s(x37)), s(0))_>=_TEST(s(s(x37)), s(s(0))))

(10)    (div(x40, x39, x38)=false & s(s(x39))=x38 & (div(x40, x39, x38)=false & s(x39)=x38  ==>  IF2(false, s(x40), s(x39))_>=_TEST(s(x40), s(s(x39))))  ==>  IF2(false, s(s(x40)), s(s(x39)))_>=_TEST(s(s(x40)), s(s(s(x39)))))



We simplified constraint (8) using rules (I), (II), (IV) which results in the following new constraint:

(11)    (IF2(false, s(0), s(s(x35)))_>=_TEST(s(0), s(s(s(x35)))))



We simplified constraint (9) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint:

(12)    (IF2(false, s(s(x37)), s(0))_>=_TEST(s(s(x37)), s(s(0))))



We simplified constraint (10) using rules (III), (IV) which results in the following new constraint:

(13)    (IF2(false, s(s(x40)), s(s(x39)))_>=_TEST(s(s(x40)), s(s(s(x39)))))








For Pair IF1(true, x, y) -> IF2(div(x, y, y), x, y) the following chains were created:
*We consider the chain TEST(x16, x17) -> IF1(gt(x16, x17), x16, x17), IF1(true, x18, x19) -> IF2(div(x18, x19, x19), x18, x19) which results in the following constraint:

(1)    (IF1(gt(x16, x17), x16, x17)=IF1(true, x18, x19)  ==>  IF1(true, x18, x19)_>=_IF2(div(x18, x19, x19), x18, x19))



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

(2)    (gt(x16, x17)=true  ==>  IF1(true, x16, x17)_>=_IF2(div(x16, x17, x17), x16, x17))



We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gt(x16, x17)=true which results in the following new constraints:

(3)    (true=true  ==>  IF1(true, s(x44), 0)_>=_IF2(div(s(x44), 0, 0), s(x44), 0))

(4)    (gt(x47, x46)=true & (gt(x47, x46)=true  ==>  IF1(true, x47, x46)_>=_IF2(div(x47, x46, x46), x47, x46))  ==>  IF1(true, s(x47), s(x46))_>=_IF2(div(s(x47), s(x46), s(x46)), s(x47), s(x46)))



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

(5)    (IF1(true, s(x44), 0)_>=_IF2(div(s(x44), 0, 0), s(x44), 0))



We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (gt(x47, x46)=true  ==>  IF1(true, x47, x46)_>=_IF2(div(x47, x46, x46), x47, x46)) with sigma = [ ] which results in the following new constraint:

(6)    (IF1(true, x47, x46)_>=_IF2(div(x47, x46, x46), x47, x46)  ==>  IF1(true, s(x47), s(x46))_>=_IF2(div(s(x47), s(x46), s(x46)), s(x47), s(x46)))








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

*TEST(x, y) -> IF1(gt(x, y), x, y)

*(TEST(x2, s(x3))_>=_IF1(gt(x2, s(x3)), x2, s(x3)))




*IF2(false, x, y) -> TEST(x, s(y))

*(IF2(false, 0, s(x27))_>=_TEST(0, s(s(x27))))


*(IF2(false, s(0), s(s(x35)))_>=_TEST(s(0), s(s(s(x35)))))


*(IF2(false, s(s(x37)), s(0))_>=_TEST(s(s(x37)), s(s(0))))


*(IF2(false, s(s(x40)), s(s(x39)))_>=_TEST(s(s(x40)), s(s(s(x39)))))




*IF1(true, x, y) -> IF2(div(x, y, y), x, y)

*(IF1(true, s(x44), 0)_>=_IF2(div(s(x44), 0, 0), s(x44), 0))


*(IF1(true, x47, x46)_>=_IF2(div(x47, x46, x46), x47, x46)  ==>  IF1(true, s(x47), s(x46))_>=_IF2(div(s(x47), s(x46), s(x46)), s(x47), s(x46)))








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(IF1(x_1, x_2, x_3)) = -1 - x_1 + x_2 - x_3
   POL(IF2(x_1, x_2, x_3)) = -1 + x_1 + x_2 - x_3
   POL(TEST(x_1, x_2)) = -1 + x_1 - x_2
   POL(c) = -2
   POL(div(x_1, x_2, x_3)) = 0
   POL(false) = 0
   POL(gt(x_1, x_2)) = 0
   POL(s(x_1)) = 1 + x_1
   POL(true) = 0


The following pairs  are in P_>:
   IF2(false, x, y) -> TEST(x, s(y))
The following pairs are in P_bound:
   IF1(true, x, y) -> IF2(div(x, y, y), x, y)
The following rules are usable:
   true -> gt(s(x), 0)
   false -> gt(0, y)
   gt(x, y) -> gt(s(x), s(y))
   div(0, 0, z) -> true
   div(0, s(x), z) -> false
   div(s(x), 0, s(z)) -> div(s(x), s(z), s(z))
   div(s(x), s(y), z) -> div(x, y, z)

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

(33)
Complex Obligation (AND)

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

(34)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TEST(x, y) -> IF1(gt(x, y), x, y)
   IF1(true, x, y) -> IF2(div(x, y, y), x, y)

The TRS R consists of the following rules:

   div(0, 0, z) -> true
   div(0, s(x), z) -> false
   div(s(x), 0, s(z)) -> div(s(x), s(z), s(z))
   div(s(x), s(y), z) -> div(x, y, z)
   gt(s(x), 0) -> true
   gt(0, y) -> false
   gt(s(x), s(y)) -> gt(x, y)

The set Q consists of the following terms:

   gt(s(x0), 0)
   gt(0, x0)
   gt(s(x0), s(x1))
   div(0, 0, x0)
   div(0, s(x0), x1)
   div(s(x0), 0, s(x1))
   div(s(x0), s(x1), x2)

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

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

(36)
TRUE

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(37)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TEST(x, y) -> IF1(gt(x, y), x, y)
   IF2(false, x, y) -> TEST(x, s(y))

The TRS R consists of the following rules:

   div(0, 0, z) -> true
   div(0, s(x), z) -> false
   div(s(x), 0, s(z)) -> div(s(x), s(z), s(z))
   div(s(x), s(y), z) -> div(x, y, z)
   gt(s(x), 0) -> true
   gt(0, y) -> false
   gt(s(x), s(y)) -> gt(x, y)

The set Q consists of the following terms:

   gt(s(x0), 0)
   gt(0, x0)
   gt(s(x0), s(x1))
   div(0, 0, x0)
   div(0, s(x0), x1)
   div(s(x0), 0, s(x1))
   div(s(x0), s(x1), x2)

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

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

(39)
TRUE