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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 NonTerminationProof (⇔, 3937 ms)
↳4 NO

(0) Obligation:

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

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

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

(4) NO