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

0 QTRS
↳1 NonTerminationProof (⇔, 847 ms)
↳2 NO

(0) Obligation:

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

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

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

(2) NO