Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

circ2(cons2(a, s), t) -> cons2(msubst2(a, t), circ2(s, t))
circ2(cons2(lift, s), cons2(a, t)) -> cons2(a, circ2(s, t))
circ2(cons2(lift, s), cons2(lift, t)) -> cons2(lift, circ2(s, t))
circ2(circ2(s, t), u) -> circ2(s, circ2(t, u))
circ2(s, id) -> s
circ2(id, s) -> s
circ2(cons2(lift, s), circ2(cons2(lift, t), u)) -> circ2(cons2(lift, circ2(s, t)), u)
subst2(a, id) -> a
msubst2(a, id) -> a
msubst2(msubst2(a, s), t) -> msubst2(a, circ2(s, t))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

circ2(cons2(a, s), t) -> cons2(msubst2(a, t), circ2(s, t))
circ2(cons2(lift, s), cons2(a, t)) -> cons2(a, circ2(s, t))
circ2(cons2(lift, s), cons2(lift, t)) -> cons2(lift, circ2(s, t))
circ2(circ2(s, t), u) -> circ2(s, circ2(t, u))
circ2(s, id) -> s
circ2(id, s) -> s
circ2(cons2(lift, s), circ2(cons2(lift, t), u)) -> circ2(cons2(lift, circ2(s, t)), u)
subst2(a, id) -> a
msubst2(a, id) -> a
msubst2(msubst2(a, s), t) -> msubst2(a, circ2(s, t))

Q is empty.

Q DP problem:
The TRS P consists of the following rules:

CIRC2(circ2(s, t), u) -> CIRC2(s, circ2(t, u))
CIRC2(cons2(a, s), t) -> MSUBST2(a, t)
MSUBST2(msubst2(a, s), t) -> MSUBST2(a, circ2(s, t))
CIRC2(circ2(s, t), u) -> CIRC2(t, u)
CIRC2(cons2(lift, s), cons2(a, t)) -> CIRC2(s, t)
CIRC2(cons2(lift, s), circ2(cons2(lift, t), u)) -> CIRC2(s, t)
CIRC2(cons2(a, s), t) -> CIRC2(s, t)
CIRC2(cons2(lift, s), circ2(cons2(lift, t), u)) -> CIRC2(cons2(lift, circ2(s, t)), u)
MSUBST2(msubst2(a, s), t) -> CIRC2(s, t)
CIRC2(cons2(lift, s), cons2(lift, t)) -> CIRC2(s, t)

The TRS R consists of the following rules:

circ2(cons2(a, s), t) -> cons2(msubst2(a, t), circ2(s, t))
circ2(cons2(lift, s), cons2(a, t)) -> cons2(a, circ2(s, t))
circ2(cons2(lift, s), cons2(lift, t)) -> cons2(lift, circ2(s, t))
circ2(circ2(s, t), u) -> circ2(s, circ2(t, u))
circ2(s, id) -> s
circ2(id, s) -> s
circ2(cons2(lift, s), circ2(cons2(lift, t), u)) -> circ2(cons2(lift, circ2(s, t)), u)
subst2(a, id) -> a
msubst2(a, id) -> a
msubst2(msubst2(a, s), t) -> msubst2(a, circ2(s, t))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
QDP

Q DP problem:
The TRS P consists of the following rules:

CIRC2(circ2(s, t), u) -> CIRC2(s, circ2(t, u))
CIRC2(cons2(a, s), t) -> MSUBST2(a, t)
MSUBST2(msubst2(a, s), t) -> MSUBST2(a, circ2(s, t))
CIRC2(circ2(s, t), u) -> CIRC2(t, u)
CIRC2(cons2(lift, s), cons2(a, t)) -> CIRC2(s, t)
CIRC2(cons2(lift, s), circ2(cons2(lift, t), u)) -> CIRC2(s, t)
CIRC2(cons2(a, s), t) -> CIRC2(s, t)
CIRC2(cons2(lift, s), circ2(cons2(lift, t), u)) -> CIRC2(cons2(lift, circ2(s, t)), u)
MSUBST2(msubst2(a, s), t) -> CIRC2(s, t)
CIRC2(cons2(lift, s), cons2(lift, t)) -> CIRC2(s, t)

The TRS R consists of the following rules:

circ2(cons2(a, s), t) -> cons2(msubst2(a, t), circ2(s, t))
circ2(cons2(lift, s), cons2(a, t)) -> cons2(a, circ2(s, t))
circ2(cons2(lift, s), cons2(lift, t)) -> cons2(lift, circ2(s, t))
circ2(circ2(s, t), u) -> circ2(s, circ2(t, u))
circ2(s, id) -> s
circ2(id, s) -> s
circ2(cons2(lift, s), circ2(cons2(lift, t), u)) -> circ2(cons2(lift, circ2(s, t)), u)
subst2(a, id) -> a
msubst2(a, id) -> a
msubst2(msubst2(a, s), t) -> msubst2(a, circ2(s, t))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.