Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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

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

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
P(x1, x2)  =  x2
p(x1, x2)  =  p(x2)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented:

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

(4) Obligation:

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

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

The TRS R consists of the following rules:

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

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