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


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MAYBE
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


Quasi decreasingness of the given CTRS could not be shown:

(0) CTRS
(1) CTRSToQTRSProof [SOUND, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) QDP
(7) NonTerminationLoopProof [COMPLETE, 0 ms]
(8) NO


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(0)
Obligation:
Conditional term rewrite system:
The TRS R consists of the following rules:

   h(d) -> c(a)
   h(d) -> c(b)
   f(k(a), k(b), x) -> f(x, x, x)

The conditional TRS C consists of the following conditional rules:

   g(x) -> k(y) <= h(x) -> d, h(x) -> c(y)


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(1) CTRSToQTRSProof (SOUND)
The conditional rules have been transormed into unconditional rules according to [CTRS,AAECCNOC].
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(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   g(x) -> U1(h(x), x)
   U1(d, x) -> U2(h(x))
   U2(c(y)) -> k(y)
   h(d) -> c(a)
   h(d) -> c(b)
   f(k(a), k(b), x) -> f(x, x, x)

Q is empty.

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(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   G(x) -> U1^1(h(x), x)
   G(x) -> H(x)
   U1^1(d, x) -> U2^1(h(x))
   U1^1(d, x) -> H(x)
   F(k(a), k(b), x) -> F(x, x, x)

The TRS R consists of the following rules:

   g(x) -> U1(h(x), x)
   U1(d, x) -> U2(h(x))
   U2(c(y)) -> k(y)
   h(d) -> c(a)
   h(d) -> c(b)
   f(k(a), k(b), x) -> f(x, x, x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.
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(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(k(a), k(b), x) -> F(x, x, x)

The TRS R consists of the following rules:

   g(x) -> U1(h(x), x)
   U1(d, x) -> U2(h(x))
   U2(c(y)) -> k(y)
   h(d) -> c(a)
   h(d) -> c(b)
   f(k(a), k(b), x) -> f(x, x, x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(7) NonTerminationLoopProof (COMPLETE)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = F(U2(h(d)), U2(h(d)), x) evaluates to  t =F(x, x, x)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
* Matcher: [ ]
* Semiunifier: [x / U2(h(d))]

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Rewriting sequence

F(U2(h(d)), U2(h(d)), U2(h(d))) -> F(U2(h(d)), U2(c(b)), U2(h(d)))
with rule h(d) -> c(b) at position [1,0] and matcher [ ]

F(U2(h(d)), U2(c(b)), U2(h(d))) -> F(U2(h(d)), k(b), U2(h(d)))
with rule U2(c(y)) -> k(y) at position [1] and matcher [y / b]

F(U2(h(d)), k(b), U2(h(d))) -> F(U2(c(a)), k(b), U2(h(d)))
with rule h(d) -> c(a) at position [0,0] and matcher [ ]

F(U2(c(a)), k(b), U2(h(d))) -> F(k(a), k(b), U2(h(d)))
with rule U2(c(y)) -> k(y) at position [0] and matcher [y / a]

F(k(a), k(b), U2(h(d))) -> F(U2(h(d)), U2(h(d)), U2(h(d)))
with rule F(k(a), k(b), x) -> F(x, x, x)

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.




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(8)
NO