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


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


NO
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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 disproven:

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 16 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 14 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) QDP
(7) UsableRulesProof [EQUIVALENT, 0 ms]
(8) QDP
(9) NonTerminationLoopProof [COMPLETE, 0 ms]
(10) NO


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

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

   f(g(a)) -> f(s(g(b)))
   f(f(x)) -> b
   g(x) -> f(g(x))

Q is empty.

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

(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(a) = 1
   POL(b) = 0
   POL(f(x_1)) = x_1
   POL(g(x_1)) = x_1
   POL(s(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   f(g(a)) -> f(s(g(b)))




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

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(f(x)) -> b
   g(x) -> f(g(x))

Q is empty.

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

(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:

   G(x) -> F(g(x))
   G(x) -> G(x)

The TRS R consists of the following rules:

   f(f(x)) -> b
   g(x) -> f(g(x))

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

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

(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   G(x) -> G(x)

The TRS R consists of the following rules:

   f(f(x)) -> b
   g(x) -> f(g(x))

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

(7) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   G(x) -> G(x)

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

(9) NonTerminationLoopProof (COMPLETE)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = G(x) evaluates to  t =G(x)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
* Matcher: [ ]
* Semiunifier: [ ]

--------------------------------------------------------------------------------
Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from G(x) to G(x).




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

(10)
NO