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

0 QTRS
↳1 NonTerminationProof (⇔, 341 ms)
↳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(b(a(x1)))))
b(x1) → x1
c(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 c b → a b b b c b

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

(2) NO