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

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(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))

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:

DIV₂(div₂(x, y), z) -> TIMES₂(y, z)
PLUS₂(s₁(x), y) -> PLUS₂(x, y)
TIMES₂(s₁(x), y) -> PLUS₂(y, times₂(x, y))
QUOT₃(x, 0, s₁(z)) -> DIV₂(x, s₁(z))
QUOT₃(s₁(x), s₁(y), z) -> QUOT₃(x, y, z)
DIV₂(div₂(x, y), z) -> DIV₂(x, times₂(y, z))
TIMES₂(s₁(x), y) -> TIMES₂(x, y)
DIV₂(x, y) -> QUOT₃(x, y, y)

The TRS R consists of the following rules:

plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))

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:

DIV₂(div₂(x, y), z) -> TIMES₂(y, z)
PLUS₂(s₁(x), y) -> PLUS₂(x, y)
TIMES₂(s₁(x), y) -> PLUS₂(y, times₂(x, y))
QUOT₃(x, 0, s₁(z)) -> DIV₂(x, s₁(z))
QUOT₃(s₁(x), s₁(y), z) -> QUOT₃(x, y, z)
DIV₂(div₂(x, y), z) -> DIV₂(x, times₂(y, z))
TIMES₂(s₁(x), y) -> TIMES₂(x, y)
DIV₂(x, y) -> QUOT₃(x, y, y)

The TRS R consists of the following rules:

plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))

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(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))

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 Order [17,21] with Interpretation:

POL( PLUS₂(x₁, x₂) ) = x₁

POL( s₁(x₁) ) = x₁ + 1

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(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))

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:

TIMES₂(s₁(x), y) -> TIMES₂(x, y)

The TRS R consists of the following rules:

plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))

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.

TIMES₂(s₁(x), y) -> TIMES₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( TIMES₂(x₁, x₂) ) = x₁

POL( s₁(x₁) ) = x₁ + 1

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(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))

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:

QUOT₃(x, 0, s₁(z)) -> DIV₂(x, s₁(z))
QUOT₃(s₁(x), s₁(y), z) -> QUOT₃(x, y, z)
DIV₂(div₂(x, y), z) -> DIV₂(x, times₂(y, z))
DIV₂(x, y) -> QUOT₃(x, y, y)

The TRS R consists of the following rules:

plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))

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.

QUOT₃(s₁(x), s₁(y), z) -> QUOT₃(x, y, z)

The remaining pairs can at least be oriented weakly.

QUOT₃(x, 0, s₁(z)) -> DIV₂(x, s₁(z))
DIV₂(div₂(x, y), z) -> DIV₂(x, times₂(y, z))
DIV₂(x, y) -> QUOT₃(x, y, y)

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( QUOT₃(x₁, ..., x₃) ) = x₁

POL( s₁(x₁) ) = x₁ + 1

POL( DIV₂(x₁, x₂) ) = x₁

POL( div₂(x₁, x₂) ) = x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

QUOT₃(x, 0, s₁(z)) -> DIV₂(x, s₁(z))
DIV₂(div₂(x, y), z) -> DIV₂(x, times₂(y, z))
DIV₂(x, y) -> QUOT₃(x, y, y)

The TRS R consists of the following rules:

plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))

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.

DIV₂(div₂(x, y), z) -> DIV₂(x, times₂(y, z))

The remaining pairs can at least be oriented weakly.

QUOT₃(x, 0, s₁(z)) -> DIV₂(x, s₁(z))
DIV₂(x, y) -> QUOT₃(x, y, y)

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( DIV₂(x₁, x₂) ) = x₁

POL( div₂(x₁, x₂) ) = x₁ + 1

POL( QUOT₃(x₁, ..., x₃) ) = x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

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

QUOT₃(x, 0, s₁(z)) -> DIV₂(x, s₁(z))
DIV₂(x, y) -> QUOT₃(x, y, y)

The TRS R consists of the following rules:

plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))

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.

QUOT₃(x, 0, s₁(z)) -> DIV₂(x, s₁(z))

The remaining pairs can at least be oriented weakly.

DIV₂(x, y) -> QUOT₃(x, y, y)

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( QUOT₃(x₁, ..., x₃) ) = x₂

POL( 0 ) = 1

POL( DIV₂(x₁, x₂) ) = x₂

POL( s₁(x₁) ) = 0

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

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

DIV₂(x, y) -> QUOT₃(x, y, y)

The TRS R consists of the following rules:

plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.