Termination w.r.t. Q of the given QTRS could be disproven:

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

a(a(a(x1))) → b(c(x1))
c(b(x1)) → a(a(a(a(c(x1)))))

Q is empty.

(1) NonTerminationProof (EQUIVALENT transformation)

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 c b b b c a c c a a c c a c c c

c b b b → b c b b b c a c c a a c c a c c c
by OverlapClosure OC 3

c b b b → b c b b c b c a a c c a c c c
by OverlapClosure OC 3

c b b b → b c b c b b c a c c c
by OverlapClosure OC 3

c b b b → b c b c b c b c c
by OverlapClosure OC 3

c b b b → b c a a a b c b c c
by OverlapClosure OC 2

c b → b c a c
by OverlapClosure OC 3

c b → a a a a c
by original rule (OC 1)

a a a → b c
by original rule (OC 1)

c b b → a a b c b c c
by OverlapClosure OC 3

c b b → a a a a a b c c
by OverlapClosure OC 2

c b → a a a a c
by original rule (OC 1)

c b → a b c c
by OverlapClosure OC 3

c b → a a a a c
by original rule (OC 1)

a a a → b c
by original rule (OC 1)

a a a → b c
by original rule (OC 1)

a a a → b c
by original rule (OC 1)

c b → b c a c
by OverlapClosure OC 3

c b → a a a a c
by original rule (OC 1)

a a a → b c
by original rule (OC 1)

c b b → b c b c a a c
by OverlapClosure OC 3

c b b → b c a a a a a c
by OverlapClosure OC 2

c b → b c a c
by OverlapClosure OC 3

c b → a a a a c
by original rule (OC 1)

a a a → b c
by original rule (OC 1)

c b → a a a a c
by original rule (OC 1)

a a a → b c
by original rule (OC 1)

c b → b c a c
by OverlapClosure OC 3

c b → a a a a c
by original rule (OC 1)

a a a → b c
by original rule (OC 1)

(2) NO