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

Q is empty.

(1) 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 b b b b b → a a a c b b b b b b c c c

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

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

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

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

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

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

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

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

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

a → b
by original rule (OC 1)

a → b
by original rule (OC 1)

a → b
by original rule (OC 1)

c b b → b b c
by OverlapClosure OC 3

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

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

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

a → b
by original rule (OC 1)

a → b
by original rule (OC 1)

a c →
by original rule (OC 1)

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

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

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

a → b
by original rule (OC 1)

a → b
by original rule (OC 1)

a → b
by original rule (OC 1)

(2) NO