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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 2073 ms]
(2) NO


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

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

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(2)
NO