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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) 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(b(a(x))) -> b(a(b(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(b(a(x))) -> b(a(b(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 1. This implies Q-termination of R.
The following rules were used to construct the certificate:

   a(b(a(x))) -> b(a(b(x)))

The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
7, 8, 9, 10, 11, 12

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

Those nodes are connected through the following edges:

* 7 to 9 labelled b_1(0)* 8 to 8 labelled #_1(0)* 9 to 10 labelled a_1(0)* 9 to 11 labelled b_1(1)* 10 to 8 labelled b_1(0)* 11 to 12 labelled a_1(1)* 11 to 11 labelled b_1(1)* 12 to 8 labelled b_1(1)


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(4)
YES