Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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