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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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:

   a(c(d(x))) -> c(x)
   u(b(d(d(x)))) -> b(x)
   v(a(a(x))) -> u(v(x))
   v(a(c(x))) -> u(b(d(x)))
   v(c(x)) -> b(x)
   w(a(a(x))) -> u(w(x))
   w(a(c(x))) -> u(b(d(x)))
   w(c(x)) -> b(x)

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:

   d(c(a(x))) -> c(x)
   d(d(b(u(x)))) -> b(x)
   a(a(v(x))) -> v(u(x))
   c(a(v(x))) -> d(b(u(x)))
   c(v(x)) -> b(x)
   a(a(w(x))) -> w(u(x))
   c(a(w(x))) -> d(b(u(x)))
   c(w(x)) -> b(x)

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:

   d(c(a(x))) -> c(x)
   d(d(b(u(x)))) -> b(x)
   a(a(v(x))) -> v(u(x))
   c(a(v(x))) -> d(b(u(x)))
   c(v(x)) -> b(x)
   a(a(w(x))) -> w(u(x))
   c(a(w(x))) -> d(b(u(x)))
   c(w(x)) -> b(x)

The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
2, 3, 5, 6, 7, 11, 12

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

Those nodes are connected through the following edges:

* 2 to 3 labelled c_1(0), b_1(0), b_1(1)* 2 to 5 labelled v_1(0), w_1(0)* 2 to 6 labelled d_1(0)* 2 to 11 labelled d_1(1)* 3 to 3 labelled #_1(0)* 5 to 3 labelled u_1(0)* 6 to 7 labelled b_1(0)* 7 to 3 labelled u_1(0)* 11 to 12 labelled b_1(1)* 12 to 3 labelled u_1(1)


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(4)
YES