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

(0) Obligation:

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

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

Q is empty.

(1) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(2) Obligation:

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

a(a(a(x))) → b(a(x))
c(b(x)) → a(a(a(c(c(x)))))

Q is empty.

(3) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [OPPELT08] to show that the SRS problem is infinite.

Found the self-embedding DerivationStructure:

c b b b b → b a c b b b b a c c a c c a a c c a c c

c b b b b → b a c b b b b a c c a c c a a c c a c c
by OverlapClosure OC 3

c b b b b → b a c b b b a a a c c a c c a a c c a c c
by OverlapClosure OC 2

c b → b a c c
by OverlapClosure OC 3

c b → a a a c c
by original rule (OC 1)

a a a → b a
by original rule (OC 1)

c b b b → b b b a a a c c a c c a a c c a c c
by OverlapClosure OC 3

c b b b → b b a a a a a c c a c c a a c c a c c
by OverlapClosure OC 3

c b b b → b b a a c b a c c a a c c a c c
by OverlapClosure OC 3

c b b b → b b a a c c b a a c c a c c
by OverlapClosure OC 3

c b b b → b a a a a c c b a a c c a c c
by OverlapClosure OC 3

c b b b → b a c b b a a c c a c c
by OverlapClosure OC 2

c b → b a c c
by OverlapClosure OC 3

c b → a a a c c
by original rule (OC 1)

a a a → b a
by original rule (OC 1)

c b b → b b a a c c a c c
by OverlapClosure OC 3

c b b → b a a a a c c a c c
by OverlapClosure OC 3

c b b → b a c b a c c
by OverlapClosure OC 2

c b → b a c c
by OverlapClosure OC 3

c b → a a a c c
by original rule (OC 1)

a a a → b a
by original rule (OC 1)

c b → b a c c
by OverlapClosure OC 3

c b → a a a c c
by original rule (OC 1)

a a a → b a
by original rule (OC 1)

c b → a a a c c
by original rule (OC 1)

a a a → b a
by original rule (OC 1)

c b → a a a c c
by original rule (OC 1)

a a a → b a
by original rule (OC 1)

c b → b a c c
by OverlapClosure OC 3

c b → a a a c c
by original rule (OC 1)

a a a → b a
by original rule (OC 1)

c b → a a a c c
by original rule (OC 1)

a a a → b a
by original rule (OC 1)

a a a → b a
by original rule (OC 1)

(4) NO