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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) RFCMatchBoundsTRSProof [EQUIVALENT, 0 ms]
(2) YES


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

   0(x1) -> 1(x1)
   4(5(4(5(x1)))) -> 4(4(5(5(x1))))
   5(5(5(5(5(5(4(4(4(4(4(4(x1)))))))))))) -> 2(x1)

Q is empty.

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(1) 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:

   0(x1) -> 1(x1)
   4(5(4(5(x1)))) -> 4(4(5(5(x1))))
   5(5(5(5(5(5(4(4(4(4(4(4(x1)))))))))))) -> 2(x1)

The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
1, 2, 3, 4, 5

Node 1 is start node and node 2 is final node.

Those nodes are connected through the following edges:

* 1 to 2 labelled 1_1(0), 2_1(0)* 1 to 3 labelled 4_1(0)* 2 to 2 labelled #_1(0)* 3 to 4 labelled 4_1(0)* 4 to 5 labelled 5_1(0)* 4 to 2 labelled 2_1(1)* 5 to 2 labelled 5_1(0), 2_1(1)


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(2)
YES