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:

max(L(x)) → x
max(N(L(0), L(y))) → y
max(N(L(s(x)), L(s(y)))) → s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) → max(N(L(x), L(max(N(y, z)))))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

max(L(x)) → x
max(N(L(0), L(y))) → y
max(N(L(s(x)), L(s(y)))) → s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) → max(N(L(x), L(max(N(y, z)))))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

max(L(x)) → x
max(N(L(0), L(y))) → y
max(N(L(s(x)), L(s(y)))) → s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) → max(N(L(x), L(max(N(y, z)))))

The set Q consists of the following terms:

max(L(x0))
max(N(L(0), L(x0)))
max(N(L(s(x0)), L(s(x1))))
max(N(L(x0), N(x1, x2)))

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

MAX(N(L(x), N(y, z))) → MAX(N(y, z))
MAX(N(L(x), N(y, z))) → MAX(N(L(x), L(max(N(y, z)))))
MAX(N(L(s(x)), L(s(y)))) → MAX(N(L(x), L(y)))

The TRS R consists of the following rules:

max(L(x)) → x
max(N(L(0), L(y))) → y
max(N(L(s(x)), L(s(y)))) → s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) → max(N(L(x), L(max(N(y, z)))))

The set Q consists of the following terms:

max(L(x0))
max(N(L(0), L(x0)))
max(N(L(s(x0)), L(s(x1))))
max(N(L(x0), N(x1, x2)))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

MAX(N(L(x), N(y, z))) → MAX(N(y, z))
MAX(N(L(x), N(y, z))) → MAX(N(L(x), L(max(N(y, z)))))
MAX(N(L(s(x)), L(s(y)))) → MAX(N(L(x), L(y)))

The TRS R consists of the following rules:

max(L(x)) → x
max(N(L(0), L(y))) → y
max(N(L(s(x)), L(s(y)))) → s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) → max(N(L(x), L(max(N(y, z)))))

The set Q consists of the following terms:

max(L(x0))
max(N(L(0), L(x0)))
max(N(L(s(x0)), L(s(x1))))
max(N(L(x0), N(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

MAX(N(L(x), N(y, z))) → MAX(N(y, z))
MAX(N(L(x), N(y, z))) → MAX(N(L(x), L(max(N(y, z)))))
MAX(N(L(s(x)), L(s(y)))) → MAX(N(L(x), L(y)))

The TRS R consists of the following rules:

max(L(x)) → x
max(N(L(0), L(y))) → y
max(N(L(s(x)), L(s(y)))) → s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) → max(N(L(x), L(max(N(y, z)))))

The set Q consists of the following terms:

max(L(x0))
max(N(L(0), L(x0)))
max(N(L(s(x0)), L(s(x1))))
max(N(L(x0), N(x1, x2)))

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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

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

MAX(N(L(s(x)), L(s(y)))) → MAX(N(L(x), L(y)))

The TRS R consists of the following rules:

max(L(x)) → x
max(N(L(0), L(y))) → y
max(N(L(s(x)), L(s(y)))) → s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) → max(N(L(x), L(max(N(y, z)))))

The set Q consists of the following terms:

max(L(x0))
max(N(L(0), L(x0)))
max(N(L(s(x0)), L(s(x1))))
max(N(L(x0), N(x1, x2)))

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.

MAX(N(L(s(x)), L(s(y)))) → MAX(N(L(x), L(y)))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
MAX(x1) = MAX(x1)
N(x1, x2) = N(x2)
L(x1) = L(x1)
s(x1) = s(x1)

Recursive Path Order [2].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

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

max(L(x)) → x
max(N(L(0), L(y))) → y
max(N(L(s(x)), L(s(y)))) → s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) → max(N(L(x), L(max(N(y, z)))))

The set Q consists of the following terms:

max(L(x0))
max(N(L(0), L(x0)))
max(N(L(s(x0)), L(s(x1))))
max(N(L(x0), N(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

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

MAX(N(L(x), N(y, z))) → MAX(N(y, z))

The TRS R consists of the following rules:

max(L(x)) → x
max(N(L(0), L(y))) → y
max(N(L(s(x)), L(s(y)))) → s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) → max(N(L(x), L(max(N(y, z)))))

The set Q consists of the following terms:

max(L(x0))
max(N(L(0), L(x0)))
max(N(L(s(x0)), L(s(x1))))
max(N(L(x0), N(x1, x2)))

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.

MAX(N(L(x), N(y, z))) → MAX(N(y, z))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
MAX(x1) = MAX(x1)
N(x1, x2) = N(x2)
L(x1) = x1

Recursive Path Order [2].
Precedence:

N₁ > MAX₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

max(L(x)) → x
max(N(L(0), L(y))) → y
max(N(L(s(x)), L(s(y)))) → s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) → max(N(L(x), L(max(N(y, z)))))

The set Q consists of the following terms:

max(L(x0))
max(N(L(0), L(x0)))
max(N(L(s(x0)), L(s(x1))))
max(N(L(x0), N(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.