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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) RFCMatchBoundsTRSProof [EQUIVALENT, 0 ms]
(2) 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(X)) -> X

The set Q consists of the following terms:

   f(0)
   f(s(0))
   p(s(x0))


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(1) 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:

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

The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10

Node 1 is start node and node 2 is final node.

Those nodes are connected through the following edges:

* 1 to 3 labelled cons_1(0)* 1 to 4 labelled f_1(0)* 1 to 2 labelled f_1(0), 0(0), cons_1(0), s_1(0), p_1(0), f_1(1), 0(1), cons_1(1), s_1(1), p_1(1)* 1 to 7 labelled cons_1(1)* 1 to 8 labelled f_1(1), cons_1(2)* 1 to 1 labelled cons_1(1)* 2 to 2 labelled #_1(0)* 3 to 2 labelled 0(0)* 4 to 5 labelled p_1(0)* 4 to 2 labelled 0(1)* 5 to 6 labelled s_1(0)* 6 to 2 labelled 0(0)* 7 to 2 labelled 0(1)* 8 to 9 labelled p_1(1)* 8 to 2 labelled 0(2)* 9 to 10 labelled s_1(1)* 10 to 2 labelled 0(1)


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(2)
YES