Termination w.r.t. Q of the given QTRS could be disproven:

0 QTRS
↳1 NonTerminationProof (⇔)
↳2 NO

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

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

a c b b b c c b b → c c a b a c b b b c c b b
by OverlapClosure OC 3
a c b b b c c b b → c a c b b b b c c b b
by OverlapClosure OC 3
a c b b b c c b b → c a c b b b b c c a b
by OverlapClosure OC 3
a c b b b c c b b → c a c b b b a c c a b
by OverlapClosure OC 3
a c b b b c c b b → c a c b a b a c c a b
by OverlapClosure OC 3
a c b b b c c b b → c a c a a b a c c a b
by OverlapClosure OC 2
a c b b b c c b → c a c a a b a c a c
by OverlapClosure OC 3
a c b b b c c b → c a c a b c a b a c a c
by OverlapClosure OC 3
a c b b b c c b → c a c a b a c b a c
by OverlapClosure OC 3
a c b b b c c b → c a a c b b a c
by OverlapClosure OC 2
a c b b b c → c a a a
by OverlapClosure OC 3
a c b b b c → c a b c a a
by OverlapClosure OC 2
a c b b b → c a b c a a b
by OverlapClosure OC 2
a c b → c a b a c
by original rule (OC 1)
a c b b → c a a b
by OverlapClosure OC 3
a c b b → c a b c a b
by OverlapClosure OC 2
a c b → c a b a c
by original rule (OC 1)
a c b → c a b
by OverlapClosure OC 2
a c b → c a b a c
by original rule (OC 1)
a c →
by OverlapClosure OC 2
a → b
by original rule (OC 1)
b c →
by original rule (OC 1)
b c →
by original rule (OC 1)
b c →
by original rule (OC 1)
b c →
by original rule (OC 1)
a c b → c b b a c
by OverlapClosure OC 3
a c b → c a b a c
by original rule (OC 1)
a → b
by original rule (OC 1)
a c b → c a b a c
by original rule (OC 1)
a c b → c a b a c
by original rule (OC 1)
b c →
by original rule (OC 1)
a c b → c a b
by OverlapClosure OC 2
a c b → c a b a c
by original rule (OC 1)
a c →
by OverlapClosure OC 2
a → b
by original rule (OC 1)
b 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)
a → b
by original rule (OC 1)
a c b → c a b a c
by original rule (OC 1)

(2) NO