/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


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

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:

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

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

The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
3, 4, 13, 14, 15, 16, 17, 18, 19, 20

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

Those nodes are connected through the following edges:

* 3 to 13 labelled s_1(0)* 3 to 4 labelled 0'_1(0), s_1(0), f_1(0), p_1(0), 0'_1(1), s_1(1), f_1(1), p_1(1)* 3 to 17 labelled s_1(1)* 3 to 15 labelled f_1(1)* 3 to 19 labelled f_1(2)* 4 to 4 labelled #_1(0)* 13 to 14 labelled p_1(0)* 14 to 15 labelled f_1(0)* 15 to 16 labelled f_1(0)* 16 to 4 labelled s_1(0), s_1(1), f_1(1), p_1(1), 0'_1(1)* 16 to 17 labelled s_1(1)* 16 to 19 labelled f_1(2)* 17 to 18 labelled p_1(1)* 18 to 19 labelled f_1(1)* 19 to 20 labelled f_1(1)* 20 to 4 labelled s_1(1), f_1(1), p_1(1), 0'_1(1)* 20 to 17 labelled s_1(1)* 20 to 19 labelled f_1(2)


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(4)
YES