Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/RubioSLASHquotminus.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:

plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
min₂(X, 0) -> X
min₂(s₁(X), s₁(Y)) -> min₂(X, Y)
min₂(min₂(X, Y), Z) -> min₂(X, plus₂(Y, Z))
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(min₂(X, Y), s₁(Y)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
min₂(X, 0) -> X
min₂(s₁(X), s₁(Y)) -> min₂(X, Y)
min₂(min₂(X, Y), Z) -> min₂(X, plus₂(Y, Z))
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(min₂(X, Y), s₁(Y)))

Q is empty.

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

PLUS₂(s₁(X), Y) -> PLUS₂(X, Y)
MIN₂(min₂(X, Y), Z) -> PLUS₂(Y, Z)
MIN₂(s₁(X), s₁(Y)) -> MIN₂(X, Y)
QUOT₂(s₁(X), s₁(Y)) -> MIN₂(X, Y)
MIN₂(min₂(X, Y), Z) -> MIN₂(X, plus₂(Y, Z))
QUOT₂(s₁(X), s₁(Y)) -> QUOT₂(min₂(X, Y), s₁(Y))

The TRS R consists of the following rules:

plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
min₂(X, 0) -> X
min₂(s₁(X), s₁(Y)) -> min₂(X, Y)
min₂(min₂(X, Y), Z) -> min₂(X, plus₂(Y, Z))
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(min₂(X, Y), s₁(Y)))

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:

PLUS₂(s₁(X), Y) -> PLUS₂(X, Y)
MIN₂(min₂(X, Y), Z) -> PLUS₂(Y, Z)
MIN₂(s₁(X), s₁(Y)) -> MIN₂(X, Y)
QUOT₂(s₁(X), s₁(Y)) -> MIN₂(X, Y)
MIN₂(min₂(X, Y), Z) -> MIN₂(X, plus₂(Y, Z))
QUOT₂(s₁(X), s₁(Y)) -> QUOT₂(min₂(X, Y), s₁(Y))

The TRS R consists of the following rules:

plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
min₂(X, 0) -> X
min₂(s₁(X), s₁(Y)) -> min₂(X, Y)
min₂(min₂(X, Y), Z) -> min₂(X, plus₂(Y, Z))
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(min₂(X, Y), s₁(Y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

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

PLUS₂(s₁(X), Y) -> PLUS₂(X, Y)

The TRS R consists of the following rules:

plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
min₂(X, 0) -> X
min₂(s₁(X), s₁(Y)) -> min₂(X, Y)
min₂(min₂(X, Y), Z) -> min₂(X, plus₂(Y, Z))
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(min₂(X, Y), s₁(Y)))

Q is empty.
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.

PLUS₂(s₁(X), Y) -> PLUS₂(X, Y)

Used argument filtering: PLUS₂(x1, x2) = x1
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
min₂(X, 0) -> X
min₂(s₁(X), s₁(Y)) -> min₂(X, Y)
min₂(min₂(X, Y), Z) -> min₂(X, plus₂(Y, Z))
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(min₂(X, Y), s₁(Y)))

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

            ↳ QDPAfsSolverProof

          ↳ QDP

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

MIN₂(s₁(X), s₁(Y)) -> MIN₂(X, Y)
MIN₂(min₂(X, Y), Z) -> MIN₂(X, plus₂(Y, Z))

The TRS R consists of the following rules:

plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
min₂(X, 0) -> X
min₂(s₁(X), s₁(Y)) -> min₂(X, Y)
min₂(min₂(X, Y), Z) -> min₂(X, plus₂(Y, Z))
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(min₂(X, Y), s₁(Y)))

Q is empty.
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.

MIN₂(min₂(X, Y), Z) -> MIN₂(X, plus₂(Y, Z))

Used argument filtering: MIN₂(x1, x2) = x1
s₁(x1) = x1
min₂(x1, x2) = min₁(x1)
plus₂(x1, x2) = x2
0 = 0
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

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

MIN₂(s₁(X), s₁(Y)) -> MIN₂(X, Y)

The TRS R consists of the following rules:

plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
min₂(X, 0) -> X
min₂(s₁(X), s₁(Y)) -> min₂(X, Y)
min₂(min₂(X, Y), Z) -> min₂(X, plus₂(Y, Z))
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(min₂(X, Y), s₁(Y)))

Q is empty.
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.

MIN₂(s₁(X), s₁(Y)) -> MIN₂(X, Y)

Used argument filtering: MIN₂(x1, x2) = x2
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

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

plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
min₂(X, 0) -> X
min₂(s₁(X), s₁(Y)) -> min₂(X, Y)
min₂(min₂(X, Y), Z) -> min₂(X, plus₂(Y, Z))
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(min₂(X, Y), s₁(Y)))

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

          ↳ QDP

            ↳ QDPAfsSolverProof

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

QUOT₂(s₁(X), s₁(Y)) -> QUOT₂(min₂(X, Y), s₁(Y))

The TRS R consists of the following rules:

plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
min₂(X, 0) -> X
min₂(s₁(X), s₁(Y)) -> min₂(X, Y)
min₂(min₂(X, Y), Z) -> min₂(X, plus₂(Y, Z))
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(min₂(X, Y), s₁(Y)))

Q is empty.
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.

QUOT₂(s₁(X), s₁(Y)) -> QUOT₂(min₂(X, Y), s₁(Y))

Used argument filtering: QUOT₂(x1, x2) = x1
s₁(x1) = s₁(x1)
min₂(x1, x2) = x1
plus₂(x1, x2) = plus₂(x1, x2)
0 = 0
Used ordering: Precedence:

plus₂ > s₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

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

plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
min₂(X, 0) -> X
min₂(s₁(X), s₁(Y)) -> min₂(X, Y)
min₂(min₂(X, Y), Z) -> min₂(X, plus₂(Y, Z))
quot₂(0, s₁(Y)) -> 0
quot₂(s₁(X), s₁(Y)) -> s₁(quot₂(min₂(X, Y), s₁(Y)))

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.