Termination w.r.t. Q proof of TRS/AG01#3.16.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:

times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(0, x) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(x, y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(0, x) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(x, 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)
TIMES₂(x, s₁(y)) -> TIMES₂(x, y)
TIMES₂(x, s₁(y)) -> PLUS₂(times₂(x, y), x)
PLUS₂(x, s₁(y)) -> PLUS₂(x, y)

The TRS R consists of the following rules:

times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(0, x) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(x, 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)
TIMES₂(x, s₁(y)) -> TIMES₂(x, y)
TIMES₂(x, s₁(y)) -> PLUS₂(times₂(x, y), x)
PLUS₂(x, s₁(y)) -> PLUS₂(x, y)

The TRS R consists of the following rules:

times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(0, x) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(x, y))

Q is empty.
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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

PLUS₂(s₁(x), y) -> PLUS₂(x, y)
PLUS₂(x, s₁(y)) -> PLUS₂(x, y)

The TRS R consists of the following rules:

times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(0, x) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(x, 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(x, s₁(y)) -> PLUS₂(x, y)

The remaining pairs can at least be oriented weakly.

PLUS₂(s₁(x), y) -> PLUS₂(x, y)

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

POL( s₁(x₁) ) = 2x₁ + 2

POL( PLUS₂(x₁, x₂) ) = 2x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ 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:

times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(0, x) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(x, 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 Order [17,21] with Interpretation:

POL( s₁(x₁) ) = x₁ + 1

POL( PLUS₂(x₁, x₂) ) = 2x₁ + 3x₂ + 2

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

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

times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(0, x) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(x, 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

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

TIMES₂(x, s₁(y)) -> TIMES₂(x, y)

The TRS R consists of the following rules:

times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(0, x) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(x, 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.

TIMES₂(x, s₁(y)) -> TIMES₂(x, y)

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

POL( s₁(x₁) ) = 3x₁ + 1

POL( TIMES₂(x₁, x₂) ) = 2x₁ + 2x₂ + 2

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(0, x) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(x, 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.