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

b a a a c a → c b a a a c a b b

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

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

b a → c b b
by OverlapClosure OC 3

b a → c a b b
by OverlapClosure OC 3

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

a →
by original rule (OC 1)

a →
by original rule (OC 1)

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

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

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

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

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

b a → a c b b
by OverlapClosure OC 3

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

a →
by original rule (OC 1)

b a → a c a
by OverlapClosure OC 2

b a → a c a b
by OverlapClosure OC 2

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

b →
by original rule (OC 1)

b →
by original rule (OC 1)

c a c →
by original rule (OC 1)

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

b →
by original rule (OC 1)

c a c →
by original rule (OC 1)

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

(4) NO