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

0 QTRS
↳1 NonTerminationProof (⇔, 328 ms)
↳2 NO

(0) Obligation:

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

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

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

(2) NO