/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 0 ms]
(6) YES


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

   a(b(a(a(b(a(a(a(a(a(x1)))))))))) -> a(a(a(a(a(a(b(a(a(b(a(a(b(x1)))))))))))))

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(a(a(a(a(b(a(a(b(a(x1)))))))))) -> b(a(a(b(a(a(b(a(a(a(a(a(a(x1)))))))))))))

Q is empty.

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

   a(a(a(a(a(b(a(a(b(a(x1)))))))))) -> b(a(a(b(a(a(b(a(a(a(a(a(a(x1)))))))))))))

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

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

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

   a(a(a(a(a(b(a(a(b(x))))))))) -> b(a(a(b(a(a(b(a(a(a(a(a(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 1. This implies Q-termination of R.
The following rules were used to construct the certificate:

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

The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215

Node 192 is start node and node 193 is final node.

Those nodes are connected through the following edges:

* 192 to 194 labelled b_1(0)* 193 to 193 labelled #_1(0)* 194 to 195 labelled a_1(0)* 195 to 196 labelled a_1(0)* 196 to 197 labelled b_1(0)* 197 to 198 labelled a_1(0)* 198 to 199 labelled a_1(0)* 199 to 200 labelled b_1(0)* 200 to 201 labelled a_1(0)* 200 to 205 labelled b_1(1)* 201 to 202 labelled a_1(0)* 201 to 205 labelled b_1(1)* 202 to 203 labelled a_1(0)* 202 to 205 labelled b_1(1)* 203 to 204 labelled a_1(0)* 203 to 205 labelled b_1(1)* 204 to 193 labelled a_1(0)* 204 to 205 labelled b_1(1)* 205 to 206 labelled a_1(1)* 206 to 207 labelled a_1(1)* 207 to 208 labelled b_1(1)* 208 to 209 labelled a_1(1)* 209 to 210 labelled a_1(1)* 210 to 211 labelled b_1(1)* 211 to 212 labelled a_1(1)* 211 to 205 labelled b_1(1)* 212 to 213 labelled a_1(1)* 212 to 205 labelled b_1(1)* 213 to 214 labelled a_1(1)* 213 to 205 labelled b_1(1)* 214 to 215 labelled a_1(1)* 214 to 205 labelled b_1(1)* 215 to 193 labelled a_1(1)* 215 to 205 labelled b_1(1)


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(6)
YES