Termination w.r.t. Q proof of TRS/Rubiobn129.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(s₁(X), plus₂(Y, Z)) -> plus₂(X, plus₂(s₁(s₁(Y)), Z))
plus₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> plus₂(X1, plus₂(X3, plus₂(X2, X4)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

plus₂(s₁(X), plus₂(Y, Z)) -> plus₂(X, plus₂(s₁(s₁(Y)), Z))
plus₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> plus₂(X1, plus₂(X3, plus₂(X2, X4)))

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(X1), plus₂(X2, plus₂(X3, X4))) -> PLUS₂(X3, plus₂(X2, X4))
PLUS₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> PLUS₂(X1, plus₂(X3, plus₂(X2, X4)))
PLUS₂(s₁(X), plus₂(Y, Z)) -> PLUS₂(X, plus₂(s₁(s₁(Y)), Z))
PLUS₂(s₁(X), plus₂(Y, Z)) -> PLUS₂(s₁(s₁(Y)), Z)
PLUS₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> PLUS₂(X2, X4)

The TRS R consists of the following rules:

plus₂(s₁(X), plus₂(Y, Z)) -> plus₂(X, plus₂(s₁(s₁(Y)), Z))
plus₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> plus₂(X1, plus₂(X3, plus₂(X2, X4)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

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

PLUS₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> PLUS₂(X3, plus₂(X2, X4))
PLUS₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> PLUS₂(X1, plus₂(X3, plus₂(X2, X4)))
PLUS₂(s₁(X), plus₂(Y, Z)) -> PLUS₂(X, plus₂(s₁(s₁(Y)), Z))
PLUS₂(s₁(X), plus₂(Y, Z)) -> PLUS₂(s₁(s₁(Y)), Z)
PLUS₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> PLUS₂(X2, X4)

The TRS R consists of the following rules:

plus₂(s₁(X), plus₂(Y, Z)) -> plus₂(X, plus₂(s₁(s₁(Y)), Z))
plus₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> plus₂(X1, plus₂(X3, plus₂(X2, X4)))

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(X1), plus₂(X2, plus₂(X3, X4))) -> PLUS₂(X3, plus₂(X2, X4))
PLUS₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> PLUS₂(X2, X4)

The remaining pairs can at least be oriented weakly.

PLUS₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> PLUS₂(X1, plus₂(X3, plus₂(X2, X4)))
PLUS₂(s₁(X), plus₂(Y, Z)) -> PLUS₂(X, plus₂(s₁(s₁(Y)), Z))
PLUS₂(s₁(X), plus₂(Y, Z)) -> PLUS₂(s₁(s₁(Y)), Z)

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

POL( s₁(x₁) ) = 1

POL( plus₂(x₁, x₂) ) = x₂ + 3

POL( PLUS₂(x₁, x₂) ) = max{0, x₂ - 3}

The following usable rules [14] were oriented:

plus₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> plus₂(X1, plus₂(X3, plus₂(X2, X4)))
plus₂(s₁(X), plus₂(Y, Z)) -> plus₂(X, plus₂(s₁(s₁(Y)), Z))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

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

PLUS₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> PLUS₂(X1, plus₂(X3, plus₂(X2, X4)))
PLUS₂(s₁(X), plus₂(Y, Z)) -> PLUS₂(X, plus₂(s₁(s₁(Y)), Z))
PLUS₂(s₁(X), plus₂(Y, Z)) -> PLUS₂(s₁(s₁(Y)), Z)

The TRS R consists of the following rules:

plus₂(s₁(X), plus₂(Y, Z)) -> plus₂(X, plus₂(s₁(s₁(Y)), Z))
plus₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> plus₂(X1, plus₂(X3, plus₂(X2, X4)))

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), plus₂(Y, Z)) -> PLUS₂(s₁(s₁(Y)), Z)

The remaining pairs can at least be oriented weakly.

PLUS₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> PLUS₂(X1, plus₂(X3, plus₂(X2, X4)))
PLUS₂(s₁(X), plus₂(Y, Z)) -> PLUS₂(X, plus₂(s₁(s₁(Y)), Z))

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

POL( s₁(x₁) ) = 1

POL( plus₂(x₁, x₂) ) = x₂ + 2

POL( PLUS₂(x₁, x₂) ) = 2x₂

The following usable rules [14] were oriented:

plus₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> plus₂(X1, plus₂(X3, plus₂(X2, X4)))
plus₂(s₁(X), plus₂(Y, Z)) -> plus₂(X, plus₂(s₁(s₁(Y)), Z))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

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

PLUS₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> PLUS₂(X1, plus₂(X3, plus₂(X2, X4)))
PLUS₂(s₁(X), plus₂(Y, Z)) -> PLUS₂(X, plus₂(s₁(s₁(Y)), Z))

The TRS R consists of the following rules:

plus₂(s₁(X), plus₂(Y, Z)) -> plus₂(X, plus₂(s₁(s₁(Y)), Z))
plus₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> plus₂(X1, plus₂(X3, plus₂(X2, X4)))

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(X1), plus₂(X2, plus₂(X3, X4))) -> PLUS₂(X1, plus₂(X3, plus₂(X2, X4)))
PLUS₂(s₁(X), plus₂(Y, Z)) -> PLUS₂(X, plus₂(s₁(s₁(Y)), Z))

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

POL( s₁(x₁) ) = 2x₁ + 3

POL( plus₂(x₁, x₂) ) = max{0, x₂ - 3}

POL( PLUS₂(x₁, x₂) ) = max{0, 3x₁ - 1}

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ PisEmptyProof

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

plus₂(s₁(X), plus₂(Y, Z)) -> plus₂(X, plus₂(s₁(s₁(Y)), Z))
plus₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> plus₂(X1, plus₂(X3, plus₂(X2, X4)))

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.