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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) Strip Symbols Proof [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:

   f(f(a)) -> f(g)

Q is empty.

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

   f(f(a)) -> f(g)

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

   f(a) -> g

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(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(a) -> g

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 0. This implies Q-termination of R.
The following rules were used to construct the certificate:

   f(a) -> g

The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
3, 4

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

Those nodes are connected through the following edges:

* 3 to 4 labelled g(0)* 4 to 4 labelled #_1(0)


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(4)
YES