NO
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 42.0 s]
(2) NO


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

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

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(2)
NO