NO
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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 disproven:

(0) QTRS
(1) NonTerminationProof [COMPLETE, 0 ms]
(2) NO


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

   0(q0(0(x1))) -> 0(0(q0(x1)))
   0(q0(h(x1))) -> 0(0(q0(h(x1))))
   0(q0(1(x1))) -> 0(1(q0(x1)))
   1(q0(0(x1))) -> 0(0(q1(x1)))
   1(q0(h(x1))) -> 0(0(q1(h(x1))))
   1(q0(1(x1))) -> 0(1(q1(x1)))
   1(q1(0(x1))) -> 1(0(q1(x1)))
   1(q1(h(x1))) -> 1(0(q1(h(x1))))
   1(q1(1(x1))) -> 1(1(q1(x1)))
   0(q1(0(x1))) -> 0(0(q2(x1)))
   0(q1(h(x1))) -> 0(0(q2(h(x1))))
   0(q1(1(x1))) -> 0(1(q2(x1)))
   1(q2(0(x1))) -> 1(0(q2(x1)))
   1(q2(h(x1))) -> 1(0(q2(h(x1))))
   1(q2(1(x1))) -> 1(1(q2(x1)))
   0(q2(x1)) -> q3(1(x1))
   1(q3(x1)) -> q3(1(x1))
   0(q3(x1)) -> q4(0(x1))
   1(q4(x1)) -> q4(1(x1))
   0(q4(0(x1))) -> 1(0(q5(x1)))
   0(q4(h(x1))) -> 1(0(q5(h(x1))))
   0(q4(1(x1))) -> 1(1(q5(x1)))
   1(q5(0(x1))) -> 0(0(q1(x1)))
   1(q5(h(x1))) -> 0(0(q1(h(x1))))
   1(q5(1(x1))) -> 0(1(q1(x1)))
   h(q0(x1)) -> h(0(q0(x1)))
   h(q1(x1)) -> h(0(q1(x1)))
   h(q2(x1)) -> h(0(q2(x1)))
   h(q3(x1)) -> h(0(q3(x1)))
   h(q4(x1)) -> h(0(q4(x1)))
   h(q5(x1)) -> h(0(q5(x1)))

Q is empty.

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(1) NonTerminationProof (COMPLETE)
We used the non-termination processor [OPPELT08] to show that the SRS problem is infinite.

Found the self-embedding DerivationStructure: 
"0 q0 h -> 0 0 q0 h"
0 q0 h -> 0 0 q0 h
by original rule (OC 1)

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(2)
NO