Termination w.r.t. Q proof of /tmp/tmpemJo64/z075.srs

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:

t(f(x1)) → t(c(n(x1)))
n(f(x1)) → f(n(x1))
o(f(x1)) → f(o(x1))
n(s(x1)) → f(s(x1))
o(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(x1)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

t(f(x1)) → t(c(n(x1)))
n(f(x1)) → f(n(x1))
o(f(x1)) → f(o(x1))
n(s(x1)) → f(s(x1))
o(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(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:

C(n(x1)) → C(x1)
T(f(x1)) → C(n(x1))
C(o(x1)) → O(c(x1))
T(f(x1)) → T(c(n(x1)))
C(o(x1)) → C(x1)
C(f(x1)) → C(x1)
C(n(x1)) → N(c(x1))
O(f(x1)) → O(x1)
N(f(x1)) → N(x1)
T(f(x1)) → N(x1)

The TRS R consists of the following rules:

t(f(x1)) → t(c(n(x1)))
n(f(x1)) → f(n(x1))
o(f(x1)) → f(o(x1))
n(s(x1)) → f(s(x1))
o(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(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:

C(n(x1)) → C(x1)
T(f(x1)) → C(n(x1))
C(o(x1)) → O(c(x1))
T(f(x1)) → T(c(n(x1)))
C(o(x1)) → C(x1)
C(f(x1)) → C(x1)
C(n(x1)) → N(c(x1))
O(f(x1)) → O(x1)
N(f(x1)) → N(x1)
T(f(x1)) → N(x1)

The TRS R consists of the following rules:

t(f(x1)) → t(c(n(x1)))
n(f(x1)) → f(n(x1))
o(f(x1)) → f(o(x1))
n(s(x1)) → f(s(x1))
o(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

O(f(x1)) → O(x1)

The TRS R consists of the following rules:

t(f(x1)) → t(c(n(x1)))
n(f(x1)) → f(n(x1))
o(f(x1)) → f(o(x1))
n(s(x1)) → f(s(x1))
o(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(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.

O(f(x1)) → O(x1)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(f(x₁)) = 1 + (4)x_1
POL(O(x₁)) = (4)x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

t(f(x1)) → t(c(n(x1)))
n(f(x1)) → f(n(x1))
o(f(x1)) → f(o(x1))
n(s(x1)) → f(s(x1))
o(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(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

          ↳ QDP

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

N(f(x1)) → N(x1)

The TRS R consists of the following rules:

t(f(x1)) → t(c(n(x1)))
n(f(x1)) → f(n(x1))
o(f(x1)) → f(o(x1))
n(s(x1)) → f(s(x1))
o(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(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.

N(f(x1)) → N(x1)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(N(x₁)) = (4)x_1
POL(f(x₁)) = 1 + (4)x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

t(f(x1)) → t(c(n(x1)))
n(f(x1)) → f(n(x1))
o(f(x1)) → f(o(x1))
n(s(x1)) → f(s(x1))
o(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(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

            ↳ QDPOrderProof

          ↳ QDP

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

C(n(x1)) → C(x1)
C(o(x1)) → C(x1)
C(f(x1)) → C(x1)

The TRS R consists of the following rules:

t(f(x1)) → t(c(n(x1)))
n(f(x1)) → f(n(x1))
o(f(x1)) → f(o(x1))
n(s(x1)) → f(s(x1))
o(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(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.

C(n(x1)) → C(x1)
C(o(x1)) → C(x1)
C(f(x1)) → C(x1)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(C(x₁)) = (4)x_1
POL(n(x₁)) = 4 + (4)x_1
POL(f(x₁)) = 4 + (4)x_1
POL(o(x₁)) = 4 + (3)x_1

The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

t(f(x1)) → t(c(n(x1)))
n(f(x1)) → f(n(x1))
o(f(x1)) → f(o(x1))
n(s(x1)) → f(s(x1))
o(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(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

          ↳ QDP

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

T(f(x1)) → T(c(n(x1)))

The TRS R consists of the following rules:

t(f(x1)) → t(c(n(x1)))
n(f(x1)) → f(n(x1))
o(f(x1)) → f(o(x1))
n(s(x1)) → f(s(x1))
o(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(x1)

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