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:

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

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

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

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

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

a b → b c a
by original rule (OC 1)

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

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

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

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

a b → b c a
by original rule (OC 1)

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

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

a b → b c a
by original rule (OC 1)

a b → b c a
by original rule (OC 1)

c b → b b a
by original rule (OC 1)

c b → b b a
by original rule (OC 1)

a b → b c a
by original rule (OC 1)

a b → b c a
by original rule (OC 1)

c b → b b a
by original rule (OC 1)

c b → b b a
by original rule (OC 1)

(4) NO