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(x1) → Q(Q(p(x1)))
p(p(x1)) → q(q(x1))
p(Q(Q(x1))) → Q(Q(p(x1)))
Q(p(q(x1))) → q(p(Q(x1)))
q(q(p(x1))) → p(q(q(x1)))
q(Q(x1)) → x1
Q(q(x1)) → x1
p(P(x1)) → x1
P(p(x1)) → x1

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

P(x1) → Q(Q(p(x1)))
p(p(x1)) → q(q(x1))
p(Q(Q(x1))) → Q(Q(p(x1)))
Q(p(q(x1))) → q(p(Q(x1)))
q(q(p(x1))) → p(q(q(x1)))
q(Q(x1)) → x1
Q(q(x1)) → x1
p(P(x1)) → x1
P(p(x1)) → 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:

Q1(p(q(x1))) → Q2(p(Q(x1)))
Q2(q(p(x1))) → Q2(x1)
P1(x1) → P2(x1)
Q1(p(q(x1))) → P2(Q(x1))
P1(x1) → Q1(Q(p(x1)))
P2(p(x1)) → Q2(x1)
Q2(q(p(x1))) → P2(q(q(x1)))
P2(Q(Q(x1))) → Q1(p(x1))
P2(Q(Q(x1))) → Q1(Q(p(x1)))
P2(Q(Q(x1))) → P2(x1)
P2(p(x1)) → Q2(q(x1))
Q1(p(q(x1))) → Q1(x1)
P1(x1) → Q1(p(x1))
Q2(q(p(x1))) → Q2(q(x1))

The TRS R consists of the following rules:

P(x1) → Q(Q(p(x1)))
p(p(x1)) → q(q(x1))
p(Q(Q(x1))) → Q(Q(p(x1)))
Q(p(q(x1))) → q(p(Q(x1)))
q(q(p(x1))) → p(q(q(x1)))
q(Q(x1)) → x1
Q(q(x1)) → x1
p(P(x1)) → x1
P(p(x1)) → x1

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

Q1(p(q(x1))) → Q2(p(Q(x1)))
Q2(q(p(x1))) → Q2(x1)
P1(x1) → P2(x1)
Q1(p(q(x1))) → P2(Q(x1))
P1(x1) → Q1(Q(p(x1)))
P2(p(x1)) → Q2(x1)
Q2(q(p(x1))) → P2(q(q(x1)))
P2(Q(Q(x1))) → Q1(p(x1))
P2(Q(Q(x1))) → Q1(Q(p(x1)))
P2(Q(Q(x1))) → P2(x1)
P2(p(x1)) → Q2(q(x1))
Q1(p(q(x1))) → Q1(x1)
P1(x1) → Q1(p(x1))
Q2(q(p(x1))) → Q2(q(x1))

The TRS R consists of the following rules:

P(x1) → Q(Q(p(x1)))
p(p(x1)) → q(q(x1))
p(Q(Q(x1))) → Q(Q(p(x1)))
Q(p(q(x1))) → q(p(Q(x1)))
q(q(p(x1))) → p(q(q(x1)))
q(Q(x1)) → x1
Q(q(x1)) → x1
p(P(x1)) → x1
P(p(x1)) → x1

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.

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

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

Q2(q(p(x1))) → P2(q(q(x1)))
P2(Q(Q(x1))) → Q1(p(x1))
P2(Q(Q(x1))) → Q1(Q(p(x1)))
P2(Q(Q(x1))) → P2(x1)
Q1(p(q(x1))) → Q2(p(Q(x1)))
Q1(p(q(x1))) → Q1(x1)
Q2(q(p(x1))) → Q2(x1)
P2(p(x1)) → Q2(q(x1))
Q2(q(p(x1))) → Q2(q(x1))
Q1(p(q(x1))) → P2(Q(x1))
P2(p(x1)) → Q2(x1)

The TRS R consists of the following rules:

P(x1) → Q(Q(p(x1)))
p(p(x1)) → q(q(x1))
p(Q(Q(x1))) → Q(Q(p(x1)))
Q(p(q(x1))) → q(p(Q(x1)))
q(q(p(x1))) → p(q(q(x1)))
q(Q(x1)) → x1
Q(q(x1)) → x1
p(P(x1)) → x1
P(p(x1)) → 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.


Q1(p(q(x1))) → Q1(x1)
Q2(q(p(x1))) → Q2(x1)
P2(p(x1)) → Q2(q(x1))
Q2(q(p(x1))) → Q2(q(x1))
P2(p(x1)) → Q2(x1)
The remaining pairs can at least be oriented weakly.

Q2(q(p(x1))) → P2(q(q(x1)))
P2(Q(Q(x1))) → Q1(p(x1))
P2(Q(Q(x1))) → Q1(Q(p(x1)))
P2(Q(Q(x1))) → P2(x1)
Q1(p(q(x1))) → Q2(p(Q(x1)))
Q1(p(q(x1))) → P2(Q(x1))
Used ordering: Polynomial interpretation [25,35]:

POL(Q2(x1)) = x_1   
POL(Q(x1)) = x_1   
POL(P(x1)) = (4)x_1   
POL(q(x1)) = x_1   
POL(p(x1)) = 1 + (2)x_1   
POL(P2(x1)) = 1 + (2)x_1   
POL(Q1(x1)) = x_1   
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

p(Q(Q(x1))) → Q(Q(p(x1)))
Q(p(q(x1))) → q(p(Q(x1)))
p(p(x1)) → q(q(x1))
q(q(p(x1))) → p(q(q(x1)))
Q(q(x1)) → x1
q(Q(x1)) → x1
p(P(x1)) → x1



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

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

Q2(q(p(x1))) → P2(q(q(x1)))
P2(Q(Q(x1))) → Q1(p(x1))
P2(Q(Q(x1))) → Q1(Q(p(x1)))
Q1(p(q(x1))) → Q2(p(Q(x1)))
P2(Q(Q(x1))) → P2(x1)
Q1(p(q(x1))) → P2(Q(x1))

The TRS R consists of the following rules:

P(x1) → Q(Q(p(x1)))
p(p(x1)) → q(q(x1))
p(Q(Q(x1))) → Q(Q(p(x1)))
Q(p(q(x1))) → q(p(Q(x1)))
q(q(p(x1))) → p(q(q(x1)))
q(Q(x1)) → x1
Q(q(x1)) → x1
p(P(x1)) → x1
P(p(x1)) → 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.


Q1(p(q(x1))) → P2(Q(x1))
The remaining pairs can at least be oriented weakly.

Q2(q(p(x1))) → P2(q(q(x1)))
P2(Q(Q(x1))) → Q1(p(x1))
P2(Q(Q(x1))) → Q1(Q(p(x1)))
Q1(p(q(x1))) → Q2(p(Q(x1)))
P2(Q(Q(x1))) → P2(x1)
Used ordering: Polynomial interpretation [25,35]:

POL(Q2(x1)) = 2 + (4)x_1   
POL(Q(x1)) = x_1   
POL(P(x1)) = (2)x_1   
POL(q(x1)) = 1 + (4)x_1   
POL(p(x1)) = 1 + (4)x_1   
POL(P2(x1)) = 2 + (4)x_1   
POL(Q1(x1)) = 1 + x_1   
The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented:

p(Q(Q(x1))) → Q(Q(p(x1)))
Q(p(q(x1))) → q(p(Q(x1)))
p(p(x1)) → q(q(x1))
q(q(p(x1))) → p(q(q(x1)))
Q(q(x1)) → x1
q(Q(x1)) → x1
p(P(x1)) → x1



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

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

Q2(q(p(x1))) → P2(q(q(x1)))
P2(Q(Q(x1))) → Q1(p(x1))
P2(Q(Q(x1))) → Q1(Q(p(x1)))
P2(Q(Q(x1))) → P2(x1)
Q1(p(q(x1))) → Q2(p(Q(x1)))

The TRS R consists of the following rules:

P(x1) → Q(Q(p(x1)))
p(p(x1)) → q(q(x1))
p(Q(Q(x1))) → Q(Q(p(x1)))
Q(p(q(x1))) → q(p(Q(x1)))
q(q(p(x1))) → p(q(q(x1)))
q(Q(x1)) → x1
Q(q(x1)) → x1
p(P(x1)) → x1
P(p(x1)) → 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.


P2(Q(Q(x1))) → Q1(p(x1))
P2(Q(Q(x1))) → P2(x1)
The remaining pairs can at least be oriented weakly.

Q2(q(p(x1))) → P2(q(q(x1)))
P2(Q(Q(x1))) → Q1(Q(p(x1)))
Q1(p(q(x1))) → Q2(p(Q(x1)))
Used ordering: Polynomial interpretation [25,35]:

POL(Q2(x1)) = x_1   
POL(Q(x1)) = 1 + (4)x_1   
POL(P(x1)) = (2)x_1   
POL(q(x1)) = 1 + (4)x_1   
POL(p(x1)) = 1 + (4)x_1   
POL(P2(x1)) = x_1   
POL(Q1(x1)) = x_1   
The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented:

p(Q(Q(x1))) → Q(Q(p(x1)))
Q(p(q(x1))) → q(p(Q(x1)))
p(p(x1)) → q(q(x1))
q(q(p(x1))) → p(q(q(x1)))
Q(q(x1)) → x1
q(Q(x1)) → x1
p(P(x1)) → x1



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

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

Q2(q(p(x1))) → P2(q(q(x1)))
P2(Q(Q(x1))) → Q1(Q(p(x1)))
Q1(p(q(x1))) → Q2(p(Q(x1)))

The TRS R consists of the following rules:

P(x1) → Q(Q(p(x1)))
p(p(x1)) → q(q(x1))
p(Q(Q(x1))) → Q(Q(p(x1)))
Q(p(q(x1))) → q(p(Q(x1)))
q(q(p(x1))) → p(q(q(x1)))
q(Q(x1)) → x1
Q(q(x1)) → x1
p(P(x1)) → x1
P(p(x1)) → x1

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