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

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

(2) NO