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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

p(a(x0), p(b(x1), p(a(x2), x3))) → p(x2, p(a(a(x0)), p(b(x1), 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(b(x1), p(a(x2), x3))) → P(a(a(x0)), p(b(x1), x3))
The remaining pairs can at least be oriented weakly.

P(a(x0), p(b(x1), p(a(x2), x3))) → P(x2, p(a(a(x0)), p(b(x1), x3)))
Used ordering: Polynomial interpretation [25,35]:

POL(P(x1, x2)) = (1/4)x_2   
POL(a(x1)) = 0   
POL(p(x1, x2)) = 1/4 + x_2   
POL(b(x1)) = 0   
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

p(a(x0), p(b(x1), p(a(x2), x3))) → p(x2, p(a(a(x0)), p(b(x1), 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(b(x1), p(a(x2), x3))) → P(x2, p(a(a(x0)), p(b(x1), x3)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(P(x1, x2)) = (2)x_1 + (2)x_2   
POL(a(x1)) = 1 + (1/4)x_1   
POL(p(x1, x2)) = x_1 + (4)x_2   
POL(b(x1)) = 1/4   
The value of delta used in the strict ordering is 6/1.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ PisEmptyProof

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

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