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:

p(m, n, s(r)) → p(m, r, n)
p(m, s(n), 0) → p(0, n, m)
p(m, 0, 0) → m

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

p(m, n, s(r)) → p(m, r, n)
p(m, s(n), 0) → p(0, n, m)
p(m, 0, 0) → m

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

p(m, n, s(r)) → p(m, r, n)
p(m, s(n), 0) → p(0, n, m)
p(m, 0, 0) → m

The set Q consists of the following terms:

p(x0, x1, s(x2))
p(x0, s(x1), 0)
p(x0, 0, 0)

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

P(m, n, s(r)) → P(m, r, n)
P(m, s(n), 0) → P(0, n, m)

The TRS R consists of the following rules:

p(m, n, s(r)) → p(m, r, n)
p(m, s(n), 0) → p(0, n, m)
p(m, 0, 0) → m

The set Q consists of the following terms:

p(x0, x1, s(x2))
p(x0, s(x1), 0)
p(x0, 0, 0)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPOrderProof

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

P(m, n, s(r)) → P(m, r, n)
P(m, s(n), 0) → P(0, n, m)

The TRS R consists of the following rules:

p(m, n, s(r)) → p(m, r, n)
p(m, s(n), 0) → p(0, n, m)
p(m, 0, 0) → m

The set Q consists of the following terms:

p(x0, x1, s(x2))
p(x0, s(x1), 0)
p(x0, 0, 0)

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.

P(m, n, s(r)) → P(m, r, n)
P(m, s(n), 0) → P(0, n, m)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
P(x1, x2, x3) = P(x1, x2, x3)
s(x1) = s(x1)
0 = 0

Recursive Path Order [2].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ PisEmptyProof

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

p(m, n, s(r)) → p(m, r, n)
p(m, s(n), 0) → p(0, n, m)
p(m, 0, 0) → m

The set Q consists of the following terms:

p(x0, x1, s(x2))
p(x0, s(x1), 0)
p(x0, 0, 0)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.