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

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

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

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

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

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

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

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)) = 0   
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.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)))
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.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.0-0(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.1-0(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.0-0(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.1-1(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.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.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.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.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.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.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.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)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ ForwardInstantiation
            ↳ QDP
              ↳ SemLabProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
                        ↳ QDPOrderProof
QDP
                            ↳ PisEmptyProof

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.