Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar3/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)))

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 contains 2 SCCs with 1 less node.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: MAX₁(x1) = x1
N₂(x1, x2) = x2
L₁(x1) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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

                ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: MAX₁(x1) = x1
N₂(x1, x2) = N₁(x2)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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.