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(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(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(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.

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


Used ordering: POLO with Polynomial interpretation [25]:

POL(P(x1, x2)) = 2·x1 + x2   
POL(a(x1)) = x1   
POL(b(x1)) = x1   
POL(p(x1, x2)) = 1 + 2·x1 + x2   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ QDPOrderProof

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.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


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

POL(P(x1, x2)) = (4)x_1 + (4)x_2   
POL(a(x1)) = 4 + (4)x_1   
POL(p(x1, x2)) = (1/2)x_1 + (1/2)x_2   
POL(b(x1)) = 0   
The value of delta used in the strict ordering is 24.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ PisEmptyProof

Q DP problem:
P is empty.
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.
The TRS P is empty. Hence, there is no (P,Q,R) chain.