Termination w.r.t. Q proof of TRS/LJB01jones4.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:

p₃(m, n, s₁(r)) -> p₃(m, r, n)
p₃(m, s₁(n), 0) -> p₃(0, n, m)
p₃(m, 0, 0) -> m

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

p₃(m, n, s₁(r)) -> p₃(m, r, n)
p₃(m, s₁(n), 0) -> p₃(0, n, m)
p₃(m, 0, 0) -> m

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:

P₃(m, n, s₁(r)) -> P₃(m, r, n)
P₃(m, s₁(n), 0) -> P₃(0, n, m)

The TRS R consists of the following rules:

p₃(m, n, s₁(r)) -> p₃(m, r, n)
p₃(m, s₁(n), 0) -> p₃(0, n, m)
p₃(m, 0, 0) -> m

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:

P₃(m, n, s₁(r)) -> P₃(m, r, n)
P₃(m, s₁(n), 0) -> P₃(0, n, m)

The TRS R consists of the following rules:

p₃(m, n, s₁(r)) -> p₃(m, r, n)
p₃(m, s₁(n), 0) -> p₃(0, n, m)
p₃(m, 0, 0) -> m

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.

P₃(m, n, s₁(r)) -> P₃(m, r, n)
P₃(m, s₁(n), 0) -> P₃(0, n, m)

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

POL(0) = 2
POL(P₃(x₁, x₂, x₃)) = 2·x₁ + x₂ + x₃
POL(s₁(x₁)) = 3 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ PisEmptyProof

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

p₃(m, n, s₁(r)) -> p₃(m, r, n)
p₃(m, s₁(n), 0) -> p₃(0, n, m)
p₃(m, 0, 0) -> m

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.