Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar18/variousSLASH24.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

  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)))))

The set Q consists of the following terms:

max₁(L₁(x0))
max₁(N₂(L₁(0), L₁(x0)))
max₁(N₂(L₁(s₁(x0)), L₁(s₁(x1))))
max₁(N₂(L₁(x0), N₂(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:

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)))))

The set Q consists of the following terms:

max₁(L₁(x0))
max₁(N₂(L₁(0), L₁(x0)))
max₁(N₂(L₁(s₁(x0)), L₁(s₁(x1))))
max₁(N₂(L₁(x0), N₂(x1, x2)))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)))))

The set Q consists of the following terms:

max₁(L₁(x0))
max₁(N₂(L₁(0), L₁(x0)))
max₁(N₂(L₁(s₁(x0)), L₁(s₁(x1))))
max₁(N₂(L₁(x0), N₂(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

  ↳ Non-Overlap Check

    ↳ 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)))))

The set Q consists of the following terms:

max₁(L₁(x0))
max₁(N₂(L₁(0), L₁(x0)))
max₁(N₂(L₁(s₁(x0)), L₁(s₁(x1))))
max₁(N₂(L₁(x0), N₂(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MAX₁(x1) = MAX₁(x1)
N₂(x1, x2) = N₁(x1)
L₁(x1) = L₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)))))

The set Q consists of the following terms:

max₁(L₁(x0))
max₁(N₂(L₁(0), L₁(x0)))
max₁(N₂(L₁(s₁(x0)), L₁(s₁(x1))))
max₁(N₂(L₁(x0), N₂(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

  ↳ Non-Overlap Check

    ↳ 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)))))

The set Q consists of the following terms:

max₁(L₁(x0))
max₁(N₂(L₁(0), L₁(x0)))
max₁(N₂(L₁(s₁(x0)), L₁(s₁(x1))))
max₁(N₂(L₁(x0), N₂(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MAX₁(x1) = MAX₁(x1)
N₂(x1, x2) = N₂(x1, x2)
L₁(x1) = L

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)))))

The set Q consists of the following terms:

max₁(L₁(x0))
max₁(N₂(L₁(0), L₁(x0)))
max₁(N₂(L₁(s₁(x0)), L₁(s₁(x1))))
max₁(N₂(L₁(x0), N₂(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.