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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 NonTerminationProof (⇔, 4456 ms)
↳4 NO

(0) Obligation:

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

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

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

(4) NO