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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

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

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

POL(P(x1, x2)) = x2   
POL(a(x1)) = 0   
POL(b(x1)) = 0   
POL(p(x1, x2)) = 1 + x2   

The following usable rules [17] were oriented:

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



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

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

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

The TRS R consists of the following rules:

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

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(x0), p(a(b(x1)), x2)) → P(a(b(a(x2))), p(a(a(x1)), x2))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
P(x1, x2)  =  P(x1, x2)
a(x1)  =  a(x1)
p(x1, x2)  =  p(x1, x2)
b(x1)  =  b(x1)

Recursive path order with status [2].
Quasi-Precedence:
[P2, p2, b1] > a1

Status:
P2: [2,1]
a1: multiset
b1: [1]
p2: [2,1]


The following usable rules [17] were oriented:

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



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

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.