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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 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:
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15

Node 2 is start node and node 3 is final node.

Those nodes are connected through the following edges:

* 2 to 4 labelled q1_1(0), q0_1(0), q3_1(0)* 2 to 5 labelled y_1(0), a_1(0)* 2 to 7 labelled y_1(0), a_1(0)* 2 to 9 labelled q4_1(0)* 3 to 3 labelled #_1(0)* 4 to 3 labelled x_1(0), a_1(0), y_1(0)* 4 to 10 labelled q0_1(1), q1_1(1), q3_1(1)* 4 to 11 labelled a_1(1), y_1(1)* 4 to 13 labelled a_1(1), y_1(1)* 5 to 6 labelled a_1(0)* 5 to 11 labelled a_1(1)* 5 to 13 labelled a_1(1)* 6 to 3 labelled q2_1(0)* 7 to 8 labelled y_1(0)* 7 to 11 labelled y_1(1)* 7 to 13 labelled y_1(1)* 8 to 3 labelled q2_1(0)* 9 to 3 labelled bl_1(0)* 9 to 15 labelled q4_1(1)* 10 to 3 labelled x_1(1), a_1(1), y_1(1)* 10 to 10 labelled q0_1(1), q1_1(1), q3_1(1)* 10 to 11 labelled a_1(1), y_1(1)* 10 to 13 labelled a_1(1), y_1(1)* 11 to 12 labelled a_1(1)* 11 to 11 labelled a_1(1)* 11 to 13 labelled a_1(1)* 12 to 3 labelled q2_1(1)* 13 to 14 labelled y_1(1)* 13 to 11 labelled y_1(1)* 13 to 13 labelled y_1(1)* 14 to 3 labelled q2_1(1)* 15 to 3 labelled bl_1(1)* 15 to 15 labelled q4_1(1)


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(4)
YES