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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 NonTerminationProof (⇔, 752 ms)
↳4 NO

(0) Obligation:

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

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

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

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

(4) NO