Termination w.r.t. Q proof of TRS/various24.trs

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:

max₁(L₁(x)) -> x
max₁(N₂(L₁(0), L₁(y))) -> y
max₁(N₂(L₁(s₁(x)), L₁(s₁(y)))) -> s₁(max₁(N₂(L₁(x), L₁(y))))
max₁(N₂(L₁(x), N₂(y, z))) -> max₁(N₂(L₁(x), L₁(max₁(N₂(y, z)))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

max₁(L₁(x)) -> x
max₁(N₂(L₁(0), L₁(y))) -> y
max₁(N₂(L₁(s₁(x)), L₁(s₁(y)))) -> s₁(max₁(N₂(L₁(x), L₁(y))))
max₁(N₂(L₁(x), N₂(y, z))) -> max₁(N₂(L₁(x), L₁(max₁(N₂(y, z)))))

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:

MAX₁(N₂(L₁(x), N₂(y, z))) -> MAX₁(N₂(y, z))
MAX₁(N₂(L₁(x), N₂(y, z))) -> MAX₁(N₂(L₁(x), L₁(max₁(N₂(y, z)))))
MAX₁(N₂(L₁(s₁(x)), L₁(s₁(y)))) -> MAX₁(N₂(L₁(x), L₁(y)))

The TRS R consists of the following rules:

max₁(L₁(x)) -> x
max₁(N₂(L₁(0), L₁(y))) -> y
max₁(N₂(L₁(s₁(x)), L₁(s₁(y)))) -> s₁(max₁(N₂(L₁(x), L₁(y))))
max₁(N₂(L₁(x), N₂(y, z))) -> max₁(N₂(L₁(x), L₁(max₁(N₂(y, z)))))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

MAX₁(N₂(L₁(x), N₂(y, z))) -> MAX₁(N₂(y, z))
MAX₁(N₂(L₁(x), N₂(y, z))) -> MAX₁(N₂(L₁(x), L₁(max₁(N₂(y, z)))))
MAX₁(N₂(L₁(s₁(x)), L₁(s₁(y)))) -> MAX₁(N₂(L₁(x), L₁(y)))

The TRS R consists of the following rules:

max₁(L₁(x)) -> x
max₁(N₂(L₁(0), L₁(y))) -> y
max₁(N₂(L₁(s₁(x)), L₁(s₁(y)))) -> s₁(max₁(N₂(L₁(x), L₁(y))))
max₁(N₂(L₁(x), N₂(y, z))) -> max₁(N₂(L₁(x), L₁(max₁(N₂(y, z)))))

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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

MAX₁(N₂(L₁(s₁(x)), L₁(s₁(y)))) -> MAX₁(N₂(L₁(x), L₁(y)))

The TRS R consists of the following rules:

max₁(L₁(x)) -> x
max₁(N₂(L₁(0), L₁(y))) -> y
max₁(N₂(L₁(s₁(x)), L₁(s₁(y)))) -> s₁(max₁(N₂(L₁(x), L₁(y))))
max₁(N₂(L₁(x), N₂(y, z))) -> max₁(N₂(L₁(x), L₁(max₁(N₂(y, z)))))

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.

MAX₁(N₂(L₁(s₁(x)), L₁(s₁(y)))) -> MAX₁(N₂(L₁(x), L₁(y)))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(L₁(x₁)) = x₁
POL(MAX₁(x₁)) = 2·x₁
POL(N₂(x₁, x₂)) = x₂
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

max₁(L₁(x)) -> x
max₁(N₂(L₁(0), L₁(y))) -> y
max₁(N₂(L₁(s₁(x)), L₁(s₁(y)))) -> s₁(max₁(N₂(L₁(x), L₁(y))))
max₁(N₂(L₁(x), N₂(y, z))) -> max₁(N₂(L₁(x), L₁(max₁(N₂(y, z)))))

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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

MAX₁(N₂(L₁(x), N₂(y, z))) -> MAX₁(N₂(y, z))

The TRS R consists of the following rules:

max₁(L₁(x)) -> x
max₁(N₂(L₁(0), L₁(y))) -> y
max₁(N₂(L₁(s₁(x)), L₁(s₁(y)))) -> s₁(max₁(N₂(L₁(x), L₁(y))))
max₁(N₂(L₁(x), N₂(y, z))) -> max₁(N₂(L₁(x), L₁(max₁(N₂(y, z)))))

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.

MAX₁(N₂(L₁(x), N₂(y, z))) -> MAX₁(N₂(y, z))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(L₁(x₁)) = 1
POL(MAX₁(x₁)) = 2·x₁
POL(N₂(x₁, x₂)) = 2·x₁ + 2·x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

max₁(L₁(x)) -> x
max₁(N₂(L₁(0), L₁(y))) -> y
max₁(N₂(L₁(s₁(x)), L₁(s₁(y)))) -> s₁(max₁(N₂(L₁(x), L₁(y))))
max₁(N₂(L₁(x), N₂(y, z))) -> max₁(N₂(L₁(x), L₁(max₁(N₂(y, z)))))

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.