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

a b b b b a b b → b a b b b b a b b b a

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

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

a b b → c c a
by OverlapClosure OC 2

a b → c a
by original rule (OC 1)

a b → c a
by original rule (OC 1)

c c → b a b
by original rule (OC 1)

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

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

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

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

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

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

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

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

a b b → c c a
by OverlapClosure OC 2

a b → c a
by original rule (OC 1)

a b → c a
by original rule (OC 1)

c c → b c a
by OverlapClosure OC 2

c c → b a b
by original rule (OC 1)

a b → c a
by original rule (OC 1)

a a → b
by original rule (OC 1)

a b → c a
by original rule (OC 1)

c c → b c a
by OverlapClosure OC 2

c c → b a b
by original rule (OC 1)

a b → c a
by original rule (OC 1)

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

a b b → c c a
by OverlapClosure OC 2

a b → c a
by original rule (OC 1)

a b → c a
by original rule (OC 1)

c c → b a b
by original rule (OC 1)

a b → c a
by original rule (OC 1)

a a → b
by original rule (OC 1)

c c → b a b
by original rule (OC 1)

(2) NO