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

0 QTRS
↳1 NonTerminationProof (⇔, 4539 ms)
↳2 NO

(0) Obligation:

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

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

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

(2) NO