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:

a(p(x1)) → p(a(A(x1)))
a(A(x1)) → A(a(x1))
p(A(A(x1))) → a(p(x1))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a(p(x1)) → p(a(A(x1)))
a(A(x1)) → A(a(x1))
p(A(A(x1))) → a(p(x1))

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:

A1(p(x1)) → A1(A(x1))
P(A(A(x1))) → P(x1)
A1(p(x1)) → P(a(A(x1)))
P(A(A(x1))) → A1(p(x1))
A1(A(x1)) → A1(x1)

The TRS R consists of the following rules:

a(p(x1)) → p(a(A(x1)))
a(A(x1)) → A(a(x1))
p(A(A(x1))) → a(p(x1))

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:

A1(p(x1)) → A1(A(x1))
P(A(A(x1))) → P(x1)
A1(p(x1)) → P(a(A(x1)))
P(A(A(x1))) → A1(p(x1))
A1(A(x1)) → A1(x1)

The TRS R consists of the following rules:

a(p(x1)) → p(a(A(x1)))
a(A(x1)) → A(a(x1))
p(A(A(x1))) → a(p(x1))

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.


A1(p(x1)) → A1(A(x1))
The remaining pairs can at least be oriented weakly.

P(A(A(x1))) → P(x1)
A1(p(x1)) → P(a(A(x1)))
P(A(A(x1))) → A1(p(x1))
A1(A(x1)) → A1(x1)
Used ordering: Polynomial interpretation [25,35]:

POL(P(x1)) = 2 + (1/2)x_1   
POL(A1(x1)) = 1 + (1/4)x_1   
POL(a(x1)) = x_1   
POL(A(x1)) = x_1   
POL(p(x1)) = 4 + (2)x_1   
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

a(A(x1)) → A(a(x1))
a(p(x1)) → p(a(A(x1)))
p(A(A(x1))) → a(p(x1))



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

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

P(A(A(x1))) → P(x1)
A1(p(x1)) → P(a(A(x1)))
A1(A(x1)) → A1(x1)
P(A(A(x1))) → A1(p(x1))

The TRS R consists of the following rules:

a(p(x1)) → p(a(A(x1)))
a(A(x1)) → A(a(x1))
p(A(A(x1))) → a(p(x1))

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.


A1(A(x1)) → A1(x1)
The remaining pairs can at least be oriented weakly.

P(A(A(x1))) → P(x1)
A1(p(x1)) → P(a(A(x1)))
P(A(A(x1))) → A1(p(x1))
Used ordering: Polynomial interpretation [25,35]:

POL(P(x1)) = 0   
POL(A1(x1)) = (4)x_1   
POL(a(x1)) = (4)x_1   
POL(A(x1)) = 1/4 + (4)x_1   
POL(p(x1)) = 0   
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

a(A(x1)) → A(a(x1))
a(p(x1)) → p(a(A(x1)))
p(A(A(x1))) → a(p(x1))



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

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

P(A(A(x1))) → P(x1)
A1(p(x1)) → P(a(A(x1)))
P(A(A(x1))) → A1(p(x1))

The TRS R consists of the following rules:

a(p(x1)) → p(a(A(x1)))
a(A(x1)) → A(a(x1))
p(A(A(x1))) → a(p(x1))

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(x1))) → P(x1)
A1(p(x1)) → P(a(A(x1)))
The remaining pairs can at least be oriented weakly.

P(A(A(x1))) → A1(p(x1))
Used ordering: Polynomial interpretation [25,35]:

POL(P(x1)) = (1/4)x_1   
POL(A1(x1)) = 1/4 + x_1   
POL(a(x1)) = 1/4 + x_1   
POL(A(x1)) = 1/2 + x_1   
POL(p(x1)) = (1/4)x_1   
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:

a(A(x1)) → A(a(x1))
a(p(x1)) → p(a(A(x1)))
p(A(A(x1))) → a(p(x1))



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

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

P(A(A(x1))) → A1(p(x1))

The TRS R consists of the following rules:

a(p(x1)) → p(a(A(x1)))
a(A(x1)) → A(a(x1))
p(A(A(x1))) → a(p(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.