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


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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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) AAECC Innermost [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 1 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) QDP
(7) UsableRulesProof [EQUIVALENT, 0 ms]
(8) QDP
(9) QReductionProof [EQUIVALENT, 0 ms]
(10) QDP
(11) TransformationProof [EQUIVALENT, 0 ms]
(12) QDP
(13) DependencyGraphProof [EQUIVALENT, 0 ms]
(14) TRUE


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

   f(x, x) -> f(g(x), x)
   g(x) -> s(x)

Q is empty.

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

(1) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is 
   g(x) -> s(x)

The TRS R 2 is 
   f(x, x) -> f(g(x), x)

The signature Sigma is {f_2}
----------------------------------------

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

   f(x, x) -> f(g(x), x)
   g(x) -> s(x)

The set Q consists of the following terms:

   f(x0, x0)
   g(x0)


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

(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:

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

The TRS R consists of the following rules:

   f(x, x) -> f(g(x), x)
   g(x) -> s(x)

The set Q consists of the following terms:

   f(x0, x0)
   g(x0)

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:

   F(x, x) -> F(g(x), x)

The TRS R consists of the following rules:

   f(x, x) -> f(g(x), x)
   g(x) -> s(x)

The set Q consists of the following terms:

   f(x0, x0)
   g(x0)

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

(7) 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.
----------------------------------------

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

   F(x, x) -> F(g(x), x)

The TRS R consists of the following rules:

   g(x) -> s(x)

The set Q consists of the following terms:

   f(x0, x0)
   g(x0)

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

(9) 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].

   f(x0, x0)


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

(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(x, x) -> F(g(x), x)

The TRS R consists of the following rules:

   g(x) -> s(x)

The set Q consists of the following terms:

   g(x0)

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

(11) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule F(x, x) -> F(g(x), x) at position [0] we obtained the following new rules [LPAR04]:

   (F(x, x) -> F(s(x), x),F(x, x) -> F(s(x), x))


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

(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(x, x) -> F(s(x), x)

The TRS R consists of the following rules:

   g(x) -> s(x)

The set Q consists of the following terms:

   g(x0)

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

(13) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
----------------------------------------

(14)
TRUE