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
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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
      ↳ RuleRemovalProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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.
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:

A1(p(x1)) → A1(A(x1))


Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x1)) = x1   
POL(A1(x1)) = 2·x1   
POL(P(x1)) = 2 + 2·x1   
POL(a(x1)) = x1   
POL(p(x1)) = 1 + x1   



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

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 Order [21,25] with Interpretation:

POL( A(x1) ) = x1 + 1


POL( p(x1) ) = 1


POL( P(x1) ) = 1


POL( A1(x1) ) = x1


POL( a(x1) ) = x1



The following usable rules [17] were oriented:

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



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

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)
The remaining pairs can at least be oriented weakly.

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

POL(P(x1)) = (1/4)x_1   
POL(A1(x1)) = (1/2)x_1   
POL(a(x1)) = 1/4 + x_1   
POL(A(x1)) = 1/4 + x_1   
POL(p(x1)) = 1/4 + (1/2)x_1   
The value of delta used in the strict ordering is 1/8.
The following usable rules [17] were oriented:

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



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

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

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.


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)) = 15/4 + (1/4)x_1   
POL(A1(x1)) = x_1   
POL(a(x1)) = 1/4 + x_1   
POL(A(x1)) = 1/2 + x_1   
POL(p(x1)) = 4 + (1/4)x_1   
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ DependencyGraphProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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.
We have reversed the following QTRS:
The set of rules R is

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

The set Q is empty.
We have obtained the following QTRS:

p(a(x)) → A(a(p(x)))
A(a(x)) → a(A(x))
A(A(p(x))) → p(a(x))

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
    ↳ QTRS
  ↳ QTRS Reverse

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

p(a(x)) → A(a(p(x)))
A(a(x)) → a(A(x))
A(A(p(x))) → p(a(x))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

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

The set Q is empty.
We have obtained the following QTRS:

p(a(x)) → A(a(p(x)))
A(a(x)) → a(A(x))
A(A(p(x))) → p(a(x))

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse
    ↳ QTRS

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

p(a(x)) → A(a(p(x)))
A(a(x)) → a(A(x))
A(A(p(x))) → p(a(x))

Q is empty.