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(0(x1)) → s(s(0(s(s(p(x1))))))
p(s(0(x1))) → 0(x1)
p(s(s(x1))) → s(p(s(x1)))
f(s(x1)) → g(q(i(x1)))
g(x1) → f(p(p(x1)))
q(i(x1)) → q(s(x1))
q(s(x1)) → s(s(x1))
i(x1) → s(x1)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

p(0(x1)) → s(s(0(s(s(p(x1))))))
p(s(0(x1))) → 0(x1)
p(s(s(x1))) → s(p(s(x1)))
f(s(x1)) → g(q(i(x1)))
g(x1) → f(p(p(x1)))
q(i(x1)) → q(s(x1))
q(s(x1)) → s(s(x1))
i(x1) → s(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:

F(s(x1)) → G(q(i(x1)))
G(x1) → P(x1)
P(0(x1)) → P(x1)
F(s(x1)) → Q(i(x1))
Q(i(x1)) → Q(s(x1))
G(x1) → F(p(p(x1)))
F(s(x1)) → I(x1)
P(s(s(x1))) → P(s(x1))
G(x1) → P(p(x1))

The TRS R consists of the following rules:

p(0(x1)) → s(s(0(s(s(p(x1))))))
p(s(0(x1))) → 0(x1)
p(s(s(x1))) → s(p(s(x1)))
f(s(x1)) → g(q(i(x1)))
g(x1) → f(p(p(x1)))
q(i(x1)) → q(s(x1))
q(s(x1)) → s(s(x1))
i(x1) → s(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:

F(s(x1)) → G(q(i(x1)))
G(x1) → P(x1)
P(0(x1)) → P(x1)
F(s(x1)) → Q(i(x1))
Q(i(x1)) → Q(s(x1))
G(x1) → F(p(p(x1)))
F(s(x1)) → I(x1)
P(s(s(x1))) → P(s(x1))
G(x1) → P(p(x1))

The TRS R consists of the following rules:

p(0(x1)) → s(s(0(s(s(p(x1))))))
p(s(0(x1))) → 0(x1)
p(s(s(x1))) → s(p(s(x1)))
f(s(x1)) → g(q(i(x1)))
g(x1) → f(p(p(x1)))
q(i(x1)) → q(s(x1))
q(s(x1)) → s(s(x1))
i(x1) → s(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 5 less nodes.

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

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

P(s(s(x1))) → P(s(x1))

The TRS R consists of the following rules:

p(0(x1)) → s(s(0(s(s(p(x1))))))
p(s(0(x1))) → 0(x1)
p(s(s(x1))) → s(p(s(x1)))
f(s(x1)) → g(q(i(x1)))
g(x1) → f(p(p(x1)))
q(i(x1)) → q(s(x1))
q(s(x1)) → s(s(x1))
i(x1) → s(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(s(s(x1))) → P(s(x1))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(P(x1)) = (4)x_1   
POL(s(x1)) = 4 + (4)x_1   
The value of delta used in the strict ordering is 64.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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

p(0(x1)) → s(s(0(s(s(p(x1))))))
p(s(0(x1))) → 0(x1)
p(s(s(x1))) → s(p(s(x1)))
f(s(x1)) → g(q(i(x1)))
g(x1) → f(p(p(x1)))
q(i(x1)) → q(s(x1))
q(s(x1)) → s(s(x1))
i(x1) → s(x1)

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.

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

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

P(0(x1)) → P(x1)

The TRS R consists of the following rules:

p(0(x1)) → s(s(0(s(s(p(x1))))))
p(s(0(x1))) → 0(x1)
p(s(s(x1))) → s(p(s(x1)))
f(s(x1)) → g(q(i(x1)))
g(x1) → f(p(p(x1)))
q(i(x1)) → q(s(x1))
q(s(x1)) → s(s(x1))
i(x1) → s(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(0(x1)) → P(x1)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(P(x1)) = (2)x_1   
POL(0(x1)) = 4 + (4)x_1   
The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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

p(0(x1)) → s(s(0(s(s(p(x1))))))
p(s(0(x1))) → 0(x1)
p(s(s(x1))) → s(p(s(x1)))
f(s(x1)) → g(q(i(x1)))
g(x1) → f(p(p(x1)))
q(i(x1)) → q(s(x1))
q(s(x1)) → s(s(x1))
i(x1) → s(x1)

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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

F(s(x1)) → G(q(i(x1)))
G(x1) → F(p(p(x1)))

The TRS R consists of the following rules:

p(0(x1)) → s(s(0(s(s(p(x1))))))
p(s(0(x1))) → 0(x1)
p(s(s(x1))) → s(p(s(x1)))
f(s(x1)) → g(q(i(x1)))
g(x1) → f(p(p(x1)))
q(i(x1)) → q(s(x1))
q(s(x1)) → s(s(x1))
i(x1) → s(x1)

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