MAYBE Termination w.r.t. Q proof of ../tpdb/TRS/Endrullis/pair3swap.trs
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(a(x0)), p(x1, p(a(x2), x3))) → p(x2, p(a(a(b(x1))), p(a(a(x0)), x3)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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.

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(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(x2, 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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof

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

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(x2, 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.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

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)


Used ordering: POLO with Polynomial interpretation [25]:

POL(P(x1, x2)) = x1 + x2   
POL(a(x1)) = x1   
POL(b(x1)) = x1   
POL(p(x1, x2)) = 1 + x1 + x2   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ ForwardInstantiation

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.
By forward instantiating [14] the rule P(a(a(x0)), p(x1, p(a(x2), x3))) → P(x2, p(a(a(b(x1))), p(a(a(x0)), x3))) we obtained the following new rules:

P(a(a(x0)), p(x1, p(a(a(a(y_0))), x3))) → P(a(a(y_0)), p(a(a(b(x1))), p(a(a(x0)), x3)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ ForwardInstantiation
            ↳ QDP

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

P(a(a(x0)), p(x1, p(a(a(a(y_0))), x3))) → P(a(a(y_0)), 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.