/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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:

   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:
31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44

Node 31 is start node and node 32 is final node.

Those nodes are connected through the following edges:

* 31 to 33 labelled q1_1(0), q0_1(0), q3_1(0)* 31 to 34 labelled y_1(0), a_1(0)* 31 to 36 labelled y_1(0), a_1(0)* 31 to 38 labelled q4_1(0)* 32 to 32 labelled #_1(0)* 33 to 32 labelled x_1(0), a_1(0), y_1(0)* 33 to 39 labelled q0_1(1), q1_1(1), q3_1(1)* 33 to 40 labelled a_1(1), y_1(1)* 33 to 42 labelled a_1(1), y_1(1)* 34 to 35 labelled a_1(0)* 34 to 40 labelled a_1(1)* 34 to 42 labelled a_1(1)* 35 to 32 labelled q2_1(0)* 36 to 37 labelled y_1(0)* 36 to 40 labelled y_1(1)* 36 to 42 labelled y_1(1)* 37 to 32 labelled q2_1(0)* 38 to 32 labelled bl_1(0)* 38 to 44 labelled q4_1(1)* 39 to 32 labelled x_1(1), a_1(1), y_1(1)* 39 to 39 labelled q0_1(1), q1_1(1), q3_1(1)* 39 to 40 labelled a_1(1), y_1(1)* 39 to 42 labelled a_1(1), y_1(1)* 40 to 41 labelled a_1(1)* 40 to 40 labelled a_1(1)* 40 to 42 labelled a_1(1)* 41 to 32 labelled q2_1(1)* 42 to 43 labelled y_1(1)* 42 to 40 labelled y_1(1)* 42 to 42 labelled y_1(1)* 43 to 32 labelled q2_1(1)* 44 to 32 labelled bl_1(1)* 44 to 44 labelled q4_1(1)


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(4)
YES