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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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 [SOUND, 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:

   a__f(f(a)) -> a__f(g(f(a)))
   mark(f(X)) -> a__f(mark(X))
   mark(a) -> a
   mark(g(X)) -> g(X)
   a__f(X) -> f(X)

The set Q consists of the following terms:

   mark(f(x0))
   mark(a)
   mark(g(x0))
   a__f(x0)


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(1) QTRS Reverse (SOUND)
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:

   a'(f(a__f(x))) -> a'(f(g(a__f(x))))
   f(mark(X)) -> mark(a__f(X))
   a'(mark(x)) -> a'(x)
   g(mark(X)) -> g(X)
   a__f(X) -> 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:

   a'(f(a__f(x))) -> a'(f(g(a__f(x))))
   f(mark(X)) -> mark(a__f(X))
   a'(mark(x)) -> a'(x)
   g(mark(X)) -> g(X)
   a__f(X) -> f(X)

The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
3, 4, 5, 7, 9, 11, 12, 13

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

Those nodes are connected through the following edges:

* 3 to 5 labelled a'_1(0)* 3 to 9 labelled mark_1(0)* 3 to 4 labelled a'_1(0), g_1(0), f_1(0), a'_1(1), g_1(1)* 3 to 11 labelled a'_1(1)* 3 to 13 labelled mark_1(1)* 4 to 4 labelled #_1(0)* 5 to 7 labelled f_1(0)* 7 to 9 labelled g_1(0)* 7 to 13 labelled g_1(1), g_1(2)* 9 to 4 labelled a__f_1(0), f_1(1)* 9 to 13 labelled mark_1(1)* 11 to 12 labelled f_1(1)* 12 to 13 labelled g_1(1), g_1(2)* 13 to 4 labelled a__f_1(1), f_1(2)* 13 to 13 labelled mark_1(1)


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(4)
YES