Termination w.r.t. Q proof of TRS/SK902.21.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:

bin₂(x, 0) -> s₁(0)
bin₂(0, s₁(y)) -> 0
bin₂(s₁(x), s₁(y)) -> +₂(bin₂(x, s₁(y)), bin₂(x, y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

bin₂(x, 0) -> s₁(0)
bin₂(0, s₁(y)) -> 0
bin₂(s₁(x), s₁(y)) -> +₂(bin₂(x, s₁(y)), bin₂(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:

BIN₂(s₁(x), s₁(y)) -> BIN₂(x, y)
BIN₂(s₁(x), s₁(y)) -> BIN₂(x, s₁(y))

The TRS R consists of the following rules:

bin₂(x, 0) -> s₁(0)
bin₂(0, s₁(y)) -> 0
bin₂(s₁(x), s₁(y)) -> +₂(bin₂(x, s₁(y)), bin₂(x, y))

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:

BIN₂(s₁(x), s₁(y)) -> BIN₂(x, y)
BIN₂(s₁(x), s₁(y)) -> BIN₂(x, s₁(y))

The TRS R consists of the following rules:

bin₂(x, 0) -> s₁(0)
bin₂(0, s₁(y)) -> 0
bin₂(s₁(x), s₁(y)) -> +₂(bin₂(x, s₁(y)), bin₂(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.

BIN₂(s₁(x), s₁(y)) -> BIN₂(x, y)

The remaining pairs can at least be oriented weakly.

BIN₂(s₁(x), s₁(y)) -> BIN₂(x, s₁(y))

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

POL( BIN₂(x₁, x₂) ) = max{0, x₂ - 1}

POL( s₁(x₁) ) = x₁ + 2

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

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

BIN₂(s₁(x), s₁(y)) -> BIN₂(x, s₁(y))

The TRS R consists of the following rules:

bin₂(x, 0) -> s₁(0)
bin₂(0, s₁(y)) -> 0
bin₂(s₁(x), s₁(y)) -> +₂(bin₂(x, s₁(y)), bin₂(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.

BIN₂(s₁(x), s₁(y)) -> BIN₂(x, s₁(y))

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

POL( BIN₂(x₁, x₂) ) = max{0, x₁ - 1}

POL( s₁(x₁) ) = x₁ + 2

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ PisEmptyProof

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

bin₂(x, 0) -> s₁(0)
bin₂(0, s₁(y)) -> 0
bin₂(s₁(x), s₁(y)) -> +₂(bin₂(x, s₁(y)), bin₂(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.