Termination w.r.t. Q proof of TRS/Rubiobn122.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(plus₂(X, Y), Z) -> plus₂(X, plus₂(Y, Z))
times₂(X, s₁(Y)) -> plus₂(X, times₂(Y, X))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

plus₂(plus₂(X, Y), Z) -> plus₂(X, plus₂(Y, Z))
times₂(X, s₁(Y)) -> plus₂(X, times₂(Y, X))

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(plus₂(X, Y), Z) -> PLUS₂(X, plus₂(Y, Z))
PLUS₂(plus₂(X, Y), Z) -> PLUS₂(Y, Z)
TIMES₂(X, s₁(Y)) -> TIMES₂(Y, X)
TIMES₂(X, s₁(Y)) -> PLUS₂(X, times₂(Y, X))

The TRS R consists of the following rules:

plus₂(plus₂(X, Y), Z) -> plus₂(X, plus₂(Y, Z))
times₂(X, s₁(Y)) -> plus₂(X, times₂(Y, X))

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(plus₂(X, Y), Z) -> PLUS₂(X, plus₂(Y, Z))
PLUS₂(plus₂(X, Y), Z) -> PLUS₂(Y, Z)
TIMES₂(X, s₁(Y)) -> TIMES₂(Y, X)
TIMES₂(X, s₁(Y)) -> PLUS₂(X, times₂(Y, X))

The TRS R consists of the following rules:

plus₂(plus₂(X, Y), Z) -> plus₂(X, plus₂(Y, Z))
times₂(X, s₁(Y)) -> plus₂(X, times₂(Y, X))

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(plus₂(X, Y), Z) -> PLUS₂(X, plus₂(Y, Z))
PLUS₂(plus₂(X, Y), Z) -> PLUS₂(Y, Z)

The TRS R consists of the following rules:

plus₂(plus₂(X, Y), Z) -> plus₂(X, plus₂(Y, Z))
times₂(X, s₁(Y)) -> plus₂(X, times₂(Y, X))

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(plus₂(X, Y), Z) -> PLUS₂(X, plus₂(Y, Z))
PLUS₂(plus₂(X, Y), Z) -> PLUS₂(Y, Z)

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

POL(PLUS₂(x₁, x₂)) = 2·x₁
POL(plus₂(x₁, x₂)) = 2 + 2·x₁ + 2·x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

plus₂(plus₂(X, Y), Z) -> plus₂(X, plus₂(Y, Z))
times₂(X, s₁(Y)) -> plus₂(X, times₂(Y, X))

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(Y, X)

The TRS R consists of the following rules:

plus₂(plus₂(X, Y), Z) -> plus₂(X, plus₂(Y, Z))
times₂(X, s₁(Y)) -> plus₂(X, times₂(Y, X))

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(Y, X)

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

POL(TIMES₂(x₁, x₂)) = 2·x₁ + 2·x₂
POL(s₁(x₁)) = 1 + 2·x₁

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:

plus₂(plus₂(X, Y), Z) -> plus₂(X, plus₂(Y, Z))
times₂(X, s₁(Y)) -> plus₂(X, times₂(Y, X))

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.