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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 16.6 s]
(2) NO


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

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

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(2)
NO