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

0 QTRS
↳1 NonTerminationProof (⇔)
↳2 NO

(0) Obligation:

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

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

a b b c b b c → b b a b b c b b c
by OverlapClosure OC 3
a b b c b b c → b b a b b a c b b c
by OverlapClosure OC 2
a b b c → b b a b a
by OverlapClosure OC 3
a b b c → b b a c c a
by OverlapClosure OC 3
a b b c → b a b c a
by OverlapClosure OC 2
a b b c → b a a b
by OverlapClosure OC 3
a b b c → b a c b a b
by OverlapClosure OC 2
a b → b a c a
by original rule (OC 1)
a b c → b a b
by OverlapClosure OC 2
a b → b a c
by OverlapClosure OC 2
a b → b a c a
by original rule (OC 1)
a →
by original rule (OC 1)
c c → b
by original rule (OC 1)
c b →
by original rule (OC 1)
a b → b c a
by OverlapClosure OC 3
a b → b a c a
by original rule (OC 1)
a →
by original rule (OC 1)
a b → b a c
by OverlapClosure OC 2
a b → b a c a
by original rule (OC 1)
a →
by original rule (OC 1)
c c → b
by original rule (OC 1)
a b b c → b a c b b c
by OverlapClosure OC 3
a b b c → b a c b b a c
by OverlapClosure OC 2
a b → b a c a
by original rule (OC 1)
a b c → b b a c
by OverlapClosure OC 2
a b c → b a b
by OverlapClosure OC 2
a b → b a c
by OverlapClosure OC 2
a b → b a c a
by original rule (OC 1)
a →
by original rule (OC 1)
c c → b
by original rule (OC 1)
a b → b a c
by OverlapClosure OC 2
a b → b a c a
by original rule (OC 1)
a →
by original rule (OC 1)
a →
by original rule (OC 1)
a →
by original rule (OC 1)

(2) NO