Termination w.r.t. Q of the following Term Rewriting System could be proven:

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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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
      ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


CIRC2(cons2(a, s), t) -> MSUBST2(a, t)
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)
CIRC2(cons2(lift, s), cons2(lift, t)) -> CIRC2(s, t)
The remaining pairs can at least be oriented weakly.

CIRC2(circ2(s, t), u) -> CIRC2(s, circ2(t, u))
MSUBST2(msubst2(a, s), t) -> MSUBST2(a, circ2(s, t))
CIRC2(circ2(s, t), u) -> CIRC2(t, u)
MSUBST2(msubst2(a, s), t) -> CIRC2(s, t)
Used ordering: Polynomial interpretation [21]:

POL(CIRC2(x1, x2)) = 2 + 2·x1 + 2·x1·x2 + x2   
POL(MSUBST2(x1, x2)) = 2 + 2·x1 + 2·x1·x2 + x2   
POL(circ2(x1, x2)) = 2·x1 + 2·x1·x2 + x2   
POL(cons2(x1, x2)) = 2 + 3·x1 + x2   
POL(id) = 0   
POL(lift) = 2   
POL(msubst2(x1, x2)) = 2·x1 + 2·x1·x2 + x2   

The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
QDP
          ↳ DependencyGraphProof

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

CIRC2(circ2(s, t), u) -> CIRC2(s, circ2(t, u))
MSUBST2(msubst2(a, s), t) -> MSUBST2(a, circ2(s, t))
CIRC2(circ2(s, t), u) -> CIRC2(t, u)
MSUBST2(msubst2(a, s), 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.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
QDP
                ↳ QDPOrderProof
              ↳ QDP

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

CIRC2(circ2(s, t), u) -> CIRC2(s, circ2(t, u))
CIRC2(circ2(s, t), u) -> CIRC2(t, u)

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


CIRC2(circ2(s, t), u) -> CIRC2(s, circ2(t, u))
CIRC2(circ2(s, t), u) -> CIRC2(t, u)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(CIRC2(x1, x2)) = 3·x1   
POL(circ2(x1, x2)) = 3 + 3·x1 + 2·x1·x2 + 3·x2   
POL(cons2(x1, x2)) = 0   
POL(id) = 0   
POL(lift) = 0   
POL(msubst2(x1, x2)) = 3 + 3·x1 + 2·x1·x2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
              ↳ QDP

Q DP problem:
P is empty.
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.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
QDP
                ↳ QDPOrderProof

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

MSUBST2(msubst2(a, s), t) -> MSUBST2(a, 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MSUBST2(msubst2(a, s), t) -> MSUBST2(a, circ2(s, t))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MSUBST2(x1, x2)) = 2·x1   
POL(circ2(x1, x2)) = 0   
POL(cons2(x1, x2)) = 0   
POL(id) = 0   
POL(lift) = 0   
POL(msubst2(x1, x2)) = 3 + 3·x1 + 2·x1·x2 + 2·x2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof

Q DP problem:
P is empty.
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.
The TRS P is empty. Hence, there is no (P,Q,R) chain.