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)
b(a(b(c(x1)))) → b(c(c(a(a(a(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:

b a b c a b c c b c c → b c c b c c b c c a b c c b a b c a b c c b c c a a a

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

a → b
by original rule (OC 1)

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

a → b
by original rule (OC 1)

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

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

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

a → b
by original rule (OC 1)

a → b
by original rule (OC 1)

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

a → b
by original rule (OC 1)

a → b
by original rule (OC 1)

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

a → b
by original rule (OC 1)

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

a → b
by original rule (OC 1)

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

a → b
by original rule (OC 1)

a → b
by original rule (OC 1)

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

a → b
by original rule (OC 1)

a → b
by original rule (OC 1)

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

(2) NO