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(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 [1,13] 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(x2, p(a(a(x0)), p(b(x1), x3)))
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)

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
      ↳ EdgeDeletionProof

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)))
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)

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 deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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)))
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)

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 [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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(x2, p(a(a(x0)), p(b(x1), x3)))
P(a(x0), p(b(x1), p(a(x2), x3))) → 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 [13].


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: Combined order from the following AFS and order.
P(x1, x2)  =  x2
a(x1)  =  x1
p(x1, x2)  =  p(x2)
b(x1)  =  b(x1)

Lexicographic path order with status [19].
Quasi-Precedence:
p1 > b1

Status:
b1: [1]
p1: [1]


The following usable rules [14] 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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP

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.