Termination w.r.t. Q proof of TRS/Rubioquotminus.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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 3 SCCs with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

PLUS₂(s₁(X), Y) -> PLUS₂(X, Y)

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

POL(PLUS₂(x₁, x₂)) = x₁
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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

            ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

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

POL(0) = 0
POL(MIN₂(x₁, x₂)) = x₁
POL(Z) = 0
POL(min₂(x₁, x₂)) = 1 + x₁
POL(plus₂(x₁, x₂)) = 0
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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

            ↳ QDPOrderProof

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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

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

POL(0) = 0
POL(QUOT₂(x₁, x₂)) = x₁
POL(Z) = 0
POL(min₂(x₁, x₂)) = x₁
POL(plus₂(x₁, x₂)) = 0
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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.