Termination w.r.t. Q proof of TRS/AG01#3.15.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:

average₂(s₁(x), y) -> average₂(x, s₁(y))
average₂(x, s₁(s₁(s₁(y)))) -> s₁(average₂(s₁(x), y))
average₂(0, 0) -> 0
average₂(0, s₁(0)) -> 0
average₂(0, s₁(s₁(0))) -> s₁(0)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

average₂(s₁(x), y) -> average₂(x, s₁(y))
average₂(x, s₁(s₁(s₁(y)))) -> s₁(average₂(s₁(x), y))
average₂(0, 0) -> 0
average₂(0, s₁(0)) -> 0
average₂(0, s₁(s₁(0))) -> s₁(0)

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:

AVERAGE₂(x, s₁(s₁(s₁(y)))) -> AVERAGE₂(s₁(x), y)
AVERAGE₂(s₁(x), y) -> AVERAGE₂(x, s₁(y))

The TRS R consists of the following rules:

average₂(s₁(x), y) -> average₂(x, s₁(y))
average₂(x, s₁(s₁(s₁(y)))) -> s₁(average₂(s₁(x), y))
average₂(0, 0) -> 0
average₂(0, s₁(0)) -> 0
average₂(0, s₁(s₁(0))) -> s₁(0)

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:

AVERAGE₂(x, s₁(s₁(s₁(y)))) -> AVERAGE₂(s₁(x), y)
AVERAGE₂(s₁(x), y) -> AVERAGE₂(x, s₁(y))

The TRS R consists of the following rules:

average₂(s₁(x), y) -> average₂(x, s₁(y))
average₂(x, s₁(s₁(s₁(y)))) -> s₁(average₂(s₁(x), y))
average₂(0, 0) -> 0
average₂(0, s₁(0)) -> 0
average₂(0, s₁(s₁(0))) -> s₁(0)

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.

AVERAGE₂(x, s₁(s₁(s₁(y)))) -> AVERAGE₂(s₁(x), y)

The remaining pairs can at least be oriented weakly.

AVERAGE₂(s₁(x), y) -> AVERAGE₂(x, s₁(y))

Used ordering: Polynomial interpretation [21]:

POL(AVERAGE₂(x₁, x₂)) = x₁ + x₂
POL(s₁(x₁)) = 2 + x₁

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:

AVERAGE₂(s₁(x), y) -> AVERAGE₂(x, s₁(y))

The TRS R consists of the following rules:

average₂(s₁(x), y) -> average₂(x, s₁(y))
average₂(x, s₁(s₁(s₁(y)))) -> s₁(average₂(s₁(x), y))
average₂(0, 0) -> 0
average₂(0, s₁(0)) -> 0
average₂(0, s₁(s₁(0))) -> s₁(0)

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.

AVERAGE₂(s₁(x), y) -> AVERAGE₂(x, s₁(y))

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

POL(AVERAGE₂(x₁, x₂)) = 2·x₁ + x₂
POL(s₁(x₁)) = 2 + x₁

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:

average₂(s₁(x), y) -> average₂(x, s₁(y))
average₂(x, s₁(s₁(s₁(y)))) -> s₁(average₂(s₁(x), y))
average₂(0, 0) -> 0
average₂(0, s₁(0)) -> 0
average₂(0, s₁(s₁(0))) -> s₁(0)

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.