/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) 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:

   f(0) -> cons(0)
   f(s(0)) -> f(p(s(0)))
   p(s(0)) -> 0

The set Q consists of the following terms:

   f(0)
   f(s(0))
   p(s(0))


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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'(f(x)) -> 0'(cons(x))
   0'(s(f(x))) -> 0'(s(p(f(x))))
   0'(s(p(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 2. This implies Q-termination of R.
The following rules were used to construct the certificate:

   0'(f(x)) -> 0'(cons(x))
   0'(s(f(x))) -> 0'(s(p(f(x))))
   0'(s(p(x))) -> 0'(x)

The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
3, 4, 5, 6, 7, 8, 13, 14, 15, 16

Node 3 is start node and node 4 is final node.

Those nodes are connected through the following edges:

* 3 to 5 labelled 0'_1(0)* 3 to 6 labelled 0'_1(0)* 3 to 4 labelled 0'_1(0), 0'_1(1)* 3 to 13 labelled 0'_1(1)* 3 to 14 labelled 0'_1(1)* 3 to 8 labelled 0'_1(1)* 3 to 16 labelled 0'_1(2)* 4 to 4 labelled #_1(0)* 5 to 4 labelled cons_1(0)* 6 to 7 labelled s_1(0)* 7 to 8 labelled p_1(0)* 8 to 4 labelled f_1(0)* 13 to 4 labelled cons_1(1)* 14 to 15 labelled s_1(1)* 15 to 16 labelled p_1(1)* 16 to 4 labelled f_1(1), cons_1(2)


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(4)
YES