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

b c c c c c c → b c c c c c c b

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

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

b c c → b b
by OverlapClosure OC 3

b c c → c b b
by OverlapClosure OC 3

b c c → a a b
by OverlapClosure OC 2

b c → a b
by OverlapClosure OC 3

b c → c a b
by original rule (OC 1)

c →
by original rule (OC 1)

b c → a b
by OverlapClosure OC 3

b c → c a b
by original rule (OC 1)

c →
by original rule (OC 1)

a a → c b
by original rule (OC 1)

c →
by original rule (OC 1)

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

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

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

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

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

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

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

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

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

b c → c a b
by original rule (OC 1)

b c → a b
by OverlapClosure OC 3

b c → c a b
by original rule (OC 1)

c →
by original rule (OC 1)

a a → c b
by original rule (OC 1)

b c → c a b
by original rule (OC 1)

b c → c a b
by original rule (OC 1)

b →
by original rule (OC 1)

b c → c a
by OverlapClosure OC 2

b c → c a b
by original rule (OC 1)

b →
by original rule (OC 1)

a a → c b
by original rule (OC 1)

b c → c a b
by original rule (OC 1)

b →
by original rule (OC 1)

a a → c b
by original rule (OC 1)

(4) NO