/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 3 ms]
(4) YES


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

   q0(a(x1)) -> x(q1(x1))
   q1(a(x1)) -> a(q1(x1))
   q1(y(x1)) -> y(q1(x1))
   a(q1(b(x1))) -> q2(a(y(x1)))
   a(q2(a(x1))) -> q2(a(a(x1)))
   a(q2(y(x1))) -> q2(a(y(x1)))
   y(q1(b(x1))) -> q2(y(y(x1)))
   y(q2(a(x1))) -> q2(y(a(x1)))
   y(q2(y(x1))) -> q2(y(y(x1)))
   q2(x(x1)) -> x(q0(x1))
   q0(y(x1)) -> y(q3(x1))
   q3(y(x1)) -> y(q3(x1))
   q3(bl(x1)) -> bl(q4(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(q0(x1)) -> q1(x(x1))
   a(q1(x1)) -> q1(a(x1))
   y(q1(x1)) -> q1(y(x1))
   b(q1(a(x1))) -> y(a(q2(x1)))
   a(q2(a(x1))) -> a(a(q2(x1)))
   y(q2(a(x1))) -> y(a(q2(x1)))
   b(q1(y(x1))) -> y(y(q2(x1)))
   a(q2(y(x1))) -> a(y(q2(x1)))
   y(q2(y(x1))) -> y(y(q2(x1)))
   x(q2(x1)) -> q0(x(x1))
   y(q0(x1)) -> q3(y(x1))
   y(q3(x1)) -> q3(y(x1))
   bl(q3(x1)) -> q4(bl(x1))

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(q0(x1)) -> q1(x(x1))
   a(q1(x1)) -> q1(a(x1))
   y(q1(x1)) -> q1(y(x1))
   b(q1(a(x1))) -> y(a(q2(x1)))
   a(q2(a(x1))) -> a(a(q2(x1)))
   y(q2(a(x1))) -> y(a(q2(x1)))
   b(q1(y(x1))) -> y(y(q2(x1)))
   a(q2(y(x1))) -> a(y(q2(x1)))
   y(q2(y(x1))) -> y(y(q2(x1)))
   x(q2(x1)) -> q0(x(x1))
   y(q0(x1)) -> q3(y(x1))
   y(q3(x1)) -> q3(y(x1))
   bl(q3(x1)) -> q4(bl(x1))

The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
3, 4, 6, 9, 10, 12, 14, 22, 23, 24, 25, 26, 27, 28

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

Those nodes are connected through the following edges:

* 3 to 6 labelled q1_1(0), q0_1(0), q3_1(0)* 3 to 9 labelled y_1(0), a_1(0)* 3 to 12 labelled y_1(0), a_1(0)* 3 to 22 labelled q4_1(0)* 4 to 4 labelled #_1(0)* 6 to 4 labelled x_1(0), a_1(0), y_1(0)* 6 to 23 labelled q0_1(1), q1_1(1), q3_1(1)* 6 to 24 labelled a_1(1), y_1(1)* 6 to 26 labelled a_1(1), y_1(1)* 9 to 10 labelled a_1(0)* 9 to 24 labelled a_1(1)* 9 to 26 labelled a_1(1)* 10 to 4 labelled q2_1(0)* 12 to 14 labelled y_1(0)* 12 to 24 labelled y_1(1)* 12 to 26 labelled y_1(1)* 14 to 4 labelled q2_1(0)* 22 to 4 labelled bl_1(0)* 22 to 28 labelled q4_1(1)* 23 to 4 labelled x_1(1), a_1(1), y_1(1)* 23 to 23 labelled q0_1(1), q1_1(1), q3_1(1)* 23 to 24 labelled a_1(1), y_1(1)* 23 to 26 labelled a_1(1), y_1(1)* 24 to 25 labelled a_1(1)* 24 to 24 labelled a_1(1)* 24 to 26 labelled a_1(1)* 25 to 4 labelled q2_1(1)* 26 to 27 labelled y_1(1)* 26 to 24 labelled y_1(1)* 26 to 26 labelled y_1(1)* 27 to 4 labelled q2_1(1)* 28 to 4 labelled bl_1(1)* 28 to 28 labelled q4_1(1)


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(4)
YES