YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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(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:
43, 44, 45, 46, 47, 48, 49, 50, 51, 52

Node 43 is start node and node 44 is final node.

Those nodes are connected through the following edges:

* 43 to 45 labelled s_1(0)* 43 to 44 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)* 43 to 49 labelled s_1(1)* 43 to 47 labelled f_1(1)* 43 to 51 labelled f_1(2)* 44 to 44 labelled #_1(0)* 45 to 46 labelled p_1(0)* 46 to 47 labelled f_1(0)* 47 to 48 labelled f_1(0)* 48 to 44 labelled s_1(0), s_1(1), f_1(1), p_1(1), 0'_1(1)* 48 to 49 labelled s_1(1)* 48 to 51 labelled f_1(2)* 49 to 50 labelled p_1(1)* 50 to 51 labelled f_1(1)* 51 to 52 labelled f_1(1)* 52 to 44 labelled s_1(1), f_1(1), p_1(1), 0'_1(1)* 52 to 49 labelled s_1(1)* 52 to 51 labelled f_1(2)


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(4)
YES