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

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

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

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

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

a → b c
by original rule (OC 1)

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 b a b b b c
by OverlapClosure OC 3

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

a → b c
by original rule (OC 1)

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

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

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

a → b c
by original rule (OC 1)

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

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

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

a →
by original rule (OC 1)

a → b c
by original rule (OC 1)

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

a → b c
by original rule (OC 1)

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

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

a →
by original rule (OC 1)

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

a → b c
by original rule (OC 1)

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

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

a → b c
by original rule (OC 1)

a →
by original rule (OC 1)

(2) NO