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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) Strip Symbols Proof [SOUND, 0 ms]
(4) QTRS
(5) RFCMatchBoundsTRSProof [EQUIVALENT, 1 ms]
(6) YES


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(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(f(x)) -> g(f(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:

   f(f(x)) -> f(g(x))

Q is empty.

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(3) Strip Symbols Proof (SOUND)
We were given the following TRS:

   f(f(x)) -> f(g(x))

By stripping symbols from the only rule of the system, we obtained the following TRS [ENDRULLIS]: 

   f(x) -> g(x)

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(4)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(x) -> g(x)

Q is empty.

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(5) RFCMatchBoundsTRSProof (EQUIVALENT)
Termination of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 0. This implies Q-termination of R.
The following rules were used to construct the certificate:

   f(x) -> g(x)

The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
7, 8

Node 7 is start node and node 8 is final node.

Those nodes are connected through the following edges:

* 7 to 8 labelled g_1(0)* 8 to 8 labelled #_1(0)


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(6)
YES