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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 4487 ms]
(2) NO


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

   a(a(x1)) -> a(b(x1))
   b(a(c(x1))) -> c(c(a(a(a(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: 
"b a c a a c c a c c -> c c a c c a c c a a c c a b a c a a c c a c c a a a"
b a c a a c c a c c -> c c a c c a c c a a c c a b a c a a c c a c c a a a
by OverlapClosure OC 2"b a c a a c c -> c c a c c a c c a a c c a b a c a a b
by OverlapClosure OC 3"b a c a a c c -> c c a c c a c c a a b a c c a a b
by OverlapClosure OC 2"b a c a a c -> c c a c c a c c a a b a b a
by OverlapClosure OC 3"b a c a a c -> c c a c c a b a c a b a
by OverlapClosure OC 3"b a c a a c -> c c a b a c c a b a
by OverlapClosure OC 2"b a c -> c c a b a
by OverlapClosure OC 3"b a c -> c c a a a
by original rule (OC 1)""a a -> a b
by original rule (OC 1)"""a a a c -> a c c a b a
by OverlapClosure OC 2"a a -> a b
by original rule (OC 1)""b a c -> c c a b a
by OverlapClosure OC 3"b a c -> c c a a a
by original rule (OC 1)""a a -> a b
by original rule (OC 1)"""""b a c -> c c a b a
by OverlapClosure OC 3"b a c -> c c a a a
by original rule (OC 1)""a a -> a b
by original rule (OC 1)""""b a c -> c c a a b
by OverlapClosure OC 2"b a c -> c c a a a
by original rule (OC 1)""a a -> a b
by original rule (OC 1)""""b a c -> c c a a b
by OverlapClosure OC 2"b a c -> c c a a a
by original rule (OC 1)""a a -> a b
by original rule (OC 1)""""b a c -> c c a b a
by OverlapClosure OC 3"b a c -> c c a a a
by original rule (OC 1)""a a -> a b
by original rule (OC 1)""""b a c c -> c c a c c a a a
by OverlapClosure OC 2"b a c -> c c a b a
by OverlapClosure OC 3"b a c -> c c a a a
by original rule (OC 1)""a a -> a b
by original rule (OC 1)"""b a c -> c c a a a
by original rule (OC 1)""

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(2)
NO