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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

a →
by original rule (OC 1)

a →
by original rule (OC 1)

c b →
by original rule (OC 1)

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

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

a →
by original rule (OC 1)

a →
by original rule (OC 1)

c b →
by original rule (OC 1)

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

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

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

a →
by original rule (OC 1)

c b →
by original rule (OC 1)

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

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

a →
by original rule (OC 1)

a →
by original rule (OC 1)

c b →
by original rule (OC 1)

(2) NO