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

0 QTRS
↳1 NonTerminationProof (⇔, 14.4 s)
↳2 NO

(0) Obligation:

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

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

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

(2) NO