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


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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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) QTRS Reverse [EQUIVALENT, 0 ms]
(2) QTRS
(3) RFCMatchBoundsTRSProof [EQUIVALENT, 0 ms]
(4) YES


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

   g(0) -> 0
   g(s(x)) -> f(g(x))
   f(0) -> 0

Q is empty.

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(1) QTRS Reverse (EQUIVALENT)
We applied the QTRS Reverse Processor [REVERSE].
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(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   0'(g(x)) -> 0'(x)
   s(g(x)) -> g(f(x))
   0'(f(x)) -> 0'(x)

Q is empty.

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(3) RFCMatchBoundsTRSProof (EQUIVALENT)
Termination of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 1. This implies Q-termination of R.
The following rules were used to construct the certificate:

   0'(g(x)) -> 0'(x)
   s(g(x)) -> g(f(x))
   0'(f(x)) -> 0'(x)

The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
2, 4, 6

Node 2 is start node and node 4 is final node.

Those nodes are connected through the following edges:

* 2 to 4 labelled 0'_1(0), 0'_1(1)* 2 to 6 labelled g_1(0)* 4 to 4 labelled #_1(0)* 6 to 4 labelled f_1(0)


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(4)
YES