YES
proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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(0) -> s(0)
   f(s(0)) -> s(0)
   f(s(s(x))) -> f(f(s(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:

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

The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
6, 7, 8, 9, 10, 14, 15, 16, 18

Node 6 is start node and node 7 is final node.

Those nodes are connected through the following edges:

* 6 to 8 labelled 0'_1(0)* 6 to 9 labelled s_1(0)* 6 to 16 labelled 0'_1(1)* 6 to 18 labelled 0'_1(2)* 7 to 7 labelled #_1(0)* 8 to 7 labelled s_1(0)* 8 to 14 labelled s_1(1)* 9 to 10 labelled f_1(0)* 10 to 7 labelled f_1(0)* 14 to 15 labelled f_1(1)* 15 to 7 labelled f_1(1)* 16 to 7 labelled s_1(1)* 16 to 14 labelled s_1(1)* 16 to 15 labelled s_1(1)* 18 to 15 labelled s_1(2)* 18 to 7 labelled s_1(2)* 18 to 14 labelled s_1(1)


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(4)
YES