Termination w.r.t. Q proof of /tmp/tmp

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(a(x0), p(x1, p(x2, x3))) → p(x1, p(x0, p(a(x3), x3)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

p(a(x0), p(x1, p(x2, x3))) → p(x1, p(x0, p(a(x3), x3)))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

P(a(x0), p(x1, p(x2, x3))) → P(x1, p(x0, p(a(x3), x3)))
P(a(x0), p(x1, p(x2, x3))) → P(x0, p(a(x3), x3))
P(a(x0), p(x1, p(x2, x3))) → P(a(x3), x3)

The TRS R consists of the following rules:

p(a(x0), p(x1, p(x2, x3))) → p(x1, p(x0, p(a(x3), x3)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

      ↳ QDPOrderProof

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

P(a(x0), p(x1, p(x2, x3))) → P(x1, p(x0, p(a(x3), x3)))
P(a(x0), p(x1, p(x2, x3))) → P(x0, p(a(x3), x3))
P(a(x0), p(x1, p(x2, x3))) → P(a(x3), x3)

The TRS R consists of the following rules:

p(a(x0), p(x1, p(x2, x3))) → p(x1, p(x0, p(a(x3), x3)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

P(a(x0), p(x1, p(x2, x3))) → P(x0, p(a(x3), x3))
P(a(x0), p(x1, p(x2, x3))) → P(a(x3), x3)

The remaining pairs can at least be oriented weakly.

P(a(x0), p(x1, p(x2, x3))) → P(x1, p(x0, p(a(x3), x3)))

Used ordering: Polynomial interpretation [25]:

POL(P(x₁, x₂)) = x₂
POL(a(x₁)) = 0
POL(p(x₁, x₂)) = 1 + x₂

The following usable rules [17] were oriented:

p(a(x0), p(x1, p(x2, x3))) → p(x1, p(x0, p(a(x3), x3)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ ForwardInstantiation

      ↳ QDPOrderProof

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

P(a(x0), p(x1, p(x2, x3))) → P(x1, p(x0, p(a(x3), x3)))

The TRS R consists of the following rules:

p(a(x0), p(x1, p(x2, x3))) → p(x1, p(x0, p(a(x3), x3)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule P(a(x0), p(x1, p(x2, x3))) → P(x1, p(x0, p(a(x3), x3))) we obtained the following new rules:

P(a(x0), p(a(y_0), p(x2, x3))) → P(a(y_0), p(x0, p(a(x3), x3)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ ForwardInstantiation

            ↳ QDP

              ↳ SemLabProof

      ↳ QDPOrderProof

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

P(a(x0), p(a(y_0), p(x2, x3))) → P(a(y_0), p(x0, p(a(x3), x3)))

The TRS R consists of the following rules:

p(a(x0), p(x1, p(x2, x3))) → p(x1, p(x0, p(a(x3), x3)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following model for the rules of the TRS R. Interpretation over the domain with elements from 0 to 1.P: 0
a: 1
p: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:

P.1-0(a.0(x0), p.1-0(a.0(y_0), p.0-1(x2, x3))) → P.1-0(a.0(y_0), p.0-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.1(y_0), p.1-1(x2, x3))) → P.1-0(a.1(y_0), p.0-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.1(y_0), p.0-0(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.0(y_0), p.1-1(x2, x3))) → P.1-0(a.0(y_0), p.0-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.1(y_0), p.1-0(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.1(y_0), p.0-1(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.0(y_0), p.1-0(x2, x3))) → P.1-0(a.0(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.0(y_0), p.0-1(x2, x3))) → P.1-0(a.0(y_0), p.1-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.1(y_0), p.0-0(x2, x3))) → P.1-0(a.1(y_0), p.0-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.0(y_0), p.1-1(x2, x3))) → P.1-0(a.0(y_0), p.1-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.0(y_0), p.0-0(x2, x3))) → P.1-0(a.0(y_0), p.0-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.0(y_0), p.1-0(x2, x3))) → P.1-0(a.0(y_0), p.0-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.0(y_0), p.0-0(x2, x3))) → P.1-0(a.0(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.1(y_0), p.1-1(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.1(y_0), p.0-1(x2, x3))) → P.1-0(a.1(y_0), p.0-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.1(y_0), p.1-0(x2, x3))) → P.1-0(a.1(y_0), p.0-0(x0, p.1-0(a.0(x3), x3)))

The TRS R consists of the following rules:

p.1-0(a.1(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ ForwardInstantiation

            ↳ QDP

              ↳ SemLabProof

                ↳ QDP

                  ↳ DependencyGraphProof

      ↳ QDPOrderProof

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

P.1-0(a.0(x0), p.1-0(a.0(y_0), p.0-1(x2, x3))) → P.1-0(a.0(y_0), p.0-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.1(y_0), p.1-1(x2, x3))) → P.1-0(a.1(y_0), p.0-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.1(y_0), p.0-0(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.0(y_0), p.1-1(x2, x3))) → P.1-0(a.0(y_0), p.0-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.1(y_0), p.1-0(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.1(y_0), p.0-1(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.0(y_0), p.1-0(x2, x3))) → P.1-0(a.0(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.0(y_0), p.0-1(x2, x3))) → P.1-0(a.0(y_0), p.1-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.1(y_0), p.0-0(x2, x3))) → P.1-0(a.1(y_0), p.0-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.0(y_0), p.1-1(x2, x3))) → P.1-0(a.0(y_0), p.1-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.0(y_0), p.0-0(x2, x3))) → P.1-0(a.0(y_0), p.0-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.0(y_0), p.1-0(x2, x3))) → P.1-0(a.0(y_0), p.0-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.0(y_0), p.0-0(x2, x3))) → P.1-0(a.0(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.1(y_0), p.1-1(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.1(y_0), p.0-1(x2, x3))) → P.1-0(a.1(y_0), p.0-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.1(y_0), p.1-0(x2, x3))) → P.1-0(a.1(y_0), p.0-0(x0, p.1-0(a.0(x3), x3)))

The TRS R consists of the following rules:

p.1-0(a.1(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 13 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ ForwardInstantiation

            ↳ QDP

              ↳ SemLabProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                        ↳ QDPOrderProof

                      ↳ QDP

      ↳ QDPOrderProof

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

P.1-0(a.1(x0), p.1-0(a.1(y_0), p.1-1(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-1(a.1(x3), x3)))

The TRS R consists of the following rules:

p.1-0(a.1(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

P.1-0(a.1(x0), p.1-0(a.1(y_0), p.1-1(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-1(a.1(x3), x3)))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25]:

POL(P.1-0(x₁, x₂)) = x₁ + x₂
POL(a.1(x₁)) = 1 + x₁
POL(p.1-0(x₁, x₂)) = x₁
POL(p.1-1(x₁, x₂)) = 0

The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ ForwardInstantiation

            ↳ QDP

              ↳ SemLabProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

                      ↳ QDP

      ↳ QDPOrderProof

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

p.1-0(a.1(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))

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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ ForwardInstantiation

            ↳ QDP

              ↳ SemLabProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                        ↳ QDPOrderProof

      ↳ QDPOrderProof

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

P.1-0(a.1(x0), p.1-0(a.1(y_0), p.0-0(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.1(y_0), p.1-0(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))

The TRS R consists of the following rules:

p.1-0(a.1(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

P.1-0(a.1(x0), p.1-0(a.1(y_0), p.0-0(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.1(y_0), p.1-0(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25]:

POL(P.1-0(x₁, x₂)) = x₁ + x₂
POL(a.0(x₁)) = 0
POL(a.1(x₁)) = 1 + x₁
POL(p.0-0(x₁, x₂)) = x₂
POL(p.0-1(x₁, x₂)) = 1 + x₂
POL(p.1-0(x₁, x₂)) = x₁ + x₂
POL(p.1-1(x₁, x₂)) = 0

The following usable rules [17] were oriented:

p.1-0(a.0(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ ForwardInstantiation

            ↳ QDP

              ↳ SemLabProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

      ↳ QDPOrderProof

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

p.1-0(a.1(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))

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.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

P(a(x0), p(x1, p(x2, x3))) → P(a(x3), x3)

The remaining pairs can at least be oriented weakly.

P(a(x0), p(x1, p(x2, x3))) → P(x1, p(x0, p(a(x3), x3)))
P(a(x0), p(x1, p(x2, x3))) → P(x0, p(a(x3), x3))

Used ordering: Matrix interpretation [3]:
Non-tuple symbols:

M( a(x₁) ) =

/	0	\
\	0	/

/	0	0	\
\	0	0	/

x₁

M( p(x₁, x₂) ) =

/	1	\
\	0	/

/	0	0	\
\	0	0	/

x₁

/	0	1	\
\	1	1	/

x₂

Tuple symbols:

M( P(x₁, x₂) ) =

[

]

x₁

[

]

x₂

Matrix type:
We used a basic matrix type which is not further parametrizeable.

As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

p(a(x0), p(x1, p(x2, x3))) → p(x1, p(x0, p(a(x3), x3)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

      ↳ QDPOrderProof

        ↳ QDP

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

P(a(x0), p(x1, p(x2, x3))) → P(x1, p(x0, p(a(x3), x3)))
P(a(x0), p(x1, p(x2, x3))) → P(x0, p(a(x3), x3))

The TRS R consists of the following rules:

p(a(x0), p(x1, p(x2, x3))) → p(x1, p(x0, p(a(x3), x3)))

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