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

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

(2) NO