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


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YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


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

Q is empty.

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

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

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

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

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

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

The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
534, 535, 536, 537, 538, 539, 540, 541, 542, 543, 544, 545, 546, 547, 548, 549, 550, 551, 552, 553, 554, 555, 556, 557, 558, 559

Node 534 is start node and node 535 is final node.

Those nodes are connected through the following edges:

* 534 to 536 labelled b_1(0)* 535 to 535 labelled #_1(0)* 536 to 537 labelled a_1(0)* 537 to 538 labelled b_1(0)* 538 to 539 labelled a_1(0)* 539 to 540 labelled a_1(0)* 540 to 541 labelled b_1(0)* 541 to 542 labelled a_1(0)* 542 to 543 labelled b_1(0)* 543 to 544 labelled a_1(0)* 543 to 548 labelled b_1(1)* 544 to 545 labelled b_1(0)* 545 to 546 labelled a_1(0)* 545 to 548 labelled b_1(1)* 546 to 547 labelled b_1(0)* 547 to 535 labelled a_1(0)* 547 to 548 labelled b_1(1)* 548 to 549 labelled a_1(1)* 549 to 550 labelled b_1(1)* 550 to 551 labelled a_1(1)* 551 to 552 labelled a_1(1)* 552 to 553 labelled b_1(1)* 553 to 554 labelled a_1(1)* 554 to 555 labelled b_1(1)* 555 to 556 labelled a_1(1)* 555 to 548 labelled b_1(1)* 556 to 557 labelled b_1(1)* 557 to 558 labelled a_1(1)* 557 to 548 labelled b_1(1)* 558 to 559 labelled b_1(1)* 559 to 535 labelled a_1(1)* 559 to 548 labelled b_1(1)


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(6)
YES