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

0 QTRS
↳1 NonTerminationProof (⇔, 778 ms)
↳2 NO

(0) Obligation:

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

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

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

(2) NO