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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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(f(a)) -> f(g(n__f(a)))
   f(X) -> n__f(X)
   activate(n__f(X)) -> f(X)
   activate(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:

   a'(f(f(x))) -> a'(n__f(g(f(x))))
   f(X) -> n__f(X)
   n__f(activate(X)) -> f(X)
   activate(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:

   a'(f(f(x))) -> a'(n__f(g(f(x))))
   f(X) -> n__f(X)
   n__f(activate(X)) -> f(X)
   activate(X) -> X

The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
2, 10, 11, 12, 13, 14, 15, 16

Node 2 is start node and node 10 is final node.

Those nodes are connected through the following edges:

* 2 to 11 labelled a'_1(0)* 2 to 10 labelled n__f_1(0), f_1(0), a'_1(0), g_1(0), activate_1(0), n__f_1(1), f_1(1), a'_1(1), g_1(1), activate_1(1), n__f_1(2)* 2 to 14 labelled a'_1(1)* 10 to 10 labelled #_1(0)* 11 to 12 labelled n__f_1(0)* 12 to 13 labelled g_1(0)* 13 to 10 labelled f_1(0), n__f_1(1), f_1(1), n__f_1(2)* 14 to 15 labelled n__f_1(1)* 15 to 16 labelled g_1(1)* 16 to 10 labelled f_1(1), n__f_1(2)


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(4)
YES