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

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

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

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

c b a → b c c
by OverlapClosure OC 3

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

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

a →
by original rule (OC 1)

a →
by original rule (OC 1)

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

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

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

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

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

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

c b a → a a b
by OverlapClosure OC 2

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

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

c →
by original rule (OC 1)

c →
by original rule (OC 1)

a → b
by original rule (OC 1)

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

a →
by original rule (OC 1)

c →
by original rule (OC 1)

a → b
by original rule (OC 1)

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

(4) NO