Termination w.r.t. Q proof of /tmp/tmp2sENNH/z074.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:

r(r(x1)) → s(r(x1))
r(s(x1)) → s(r(x1))
r(n(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
n(u(x1)) → u(n(x1))
t(r(u(x1))) → t(c(r(x1)))
t(s(u(x1))) → t(c(r(x1)))
t(n(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(x1)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

r(r(x1)) → s(r(x1))
r(s(x1)) → s(r(x1))
r(n(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
n(u(x1)) → u(n(x1))
t(r(u(x1))) → t(c(r(x1)))
t(s(u(x1))) → t(c(r(x1)))
t(n(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(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:

R(s(x1)) → R(x1)
C(n(x1)) → N(c(x1))
T(n(u(x1))) → R(x1)
R(n(x1)) → R(x1)
R(s(x1)) → S(r(x1))
T(r(u(x1))) → T(c(r(x1)))
R(b(x1)) → S(b(x1))
C(s(x1)) → S(c(x1))
C(r(x1)) → R(c(x1))
C(s(x1)) → C(x1)
S(u(x1)) → S(x1)
R(r(x1)) → S(r(x1))
C(r(x1)) → C(x1)
C(n(x1)) → C(x1)
N(u(x1)) → N(x1)
R(n(x1)) → S(r(x1))
C(u(x1)) → C(x1)
T(r(u(x1))) → C(r(x1))
R(u(x1)) → R(x1)
T(s(u(x1))) → T(c(r(x1)))
T(n(u(x1))) → T(c(r(x1)))
T(s(u(x1))) → C(r(x1))
T(n(u(x1))) → C(r(x1))
T(s(u(x1))) → R(x1)
T(r(u(x1))) → R(x1)

The TRS R consists of the following rules:

r(r(x1)) → s(r(x1))
r(s(x1)) → s(r(x1))
r(n(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
n(u(x1)) → u(n(x1))
t(r(u(x1))) → t(c(r(x1)))
t(s(u(x1))) → t(c(r(x1)))
t(n(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(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:

R(s(x1)) → R(x1)
C(n(x1)) → N(c(x1))
T(n(u(x1))) → R(x1)
R(n(x1)) → R(x1)
R(s(x1)) → S(r(x1))
T(r(u(x1))) → T(c(r(x1)))
R(b(x1)) → S(b(x1))
C(s(x1)) → S(c(x1))
C(r(x1)) → R(c(x1))
C(s(x1)) → C(x1)
S(u(x1)) → S(x1)
R(r(x1)) → S(r(x1))
C(r(x1)) → C(x1)
C(n(x1)) → C(x1)
N(u(x1)) → N(x1)
R(n(x1)) → S(r(x1))
C(u(x1)) → C(x1)
T(r(u(x1))) → C(r(x1))
R(u(x1)) → R(x1)
T(s(u(x1))) → T(c(r(x1)))
T(n(u(x1))) → T(c(r(x1)))
T(s(u(x1))) → C(r(x1))
T(n(u(x1))) → C(r(x1))
T(s(u(x1))) → R(x1)
T(r(u(x1))) → R(x1)

The TRS R consists of the following rules:

r(r(x1)) → s(r(x1))
r(s(x1)) → s(r(x1))
r(n(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
n(u(x1)) → u(n(x1))
t(r(u(x1))) → t(c(r(x1)))
t(s(u(x1))) → t(c(r(x1)))
t(n(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 5 SCCs with 13 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

N(u(x1)) → N(x1)

The TRS R consists of the following rules:

r(r(x1)) → s(r(x1))
r(s(x1)) → s(r(x1))
r(n(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
n(u(x1)) → u(n(x1))
t(r(u(x1))) → t(c(r(x1)))
t(s(u(x1))) → t(c(r(x1)))
t(n(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(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(u(x1)) → N(x1)

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

POL(N(x₁)) = (2)x_1
POL(u(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

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

r(r(x1)) → s(r(x1))
r(s(x1)) → s(r(x1))
r(n(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
n(u(x1)) → u(n(x1))
t(r(u(x1))) → t(c(r(x1)))
t(s(u(x1))) → t(c(r(x1)))
t(n(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(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

          ↳ QDP

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

S(u(x1)) → S(x1)

The TRS R consists of the following rules:

r(r(x1)) → s(r(x1))
r(s(x1)) → s(r(x1))
r(n(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
n(u(x1)) → u(n(x1))
t(r(u(x1))) → t(c(r(x1)))
t(s(u(x1))) → t(c(r(x1)))
t(n(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(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.

S(u(x1)) → S(x1)

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

POL(u(x₁)) = 4 + (4)x_1
POL(S(x₁)) = (2)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

          ↳ QDP

          ↳ QDP

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

r(r(x1)) → s(r(x1))
r(s(x1)) → s(r(x1))
r(n(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
n(u(x1)) → u(n(x1))
t(r(u(x1))) → t(c(r(x1)))
t(s(u(x1))) → t(c(r(x1)))
t(n(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(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

          ↳ QDP

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

R(u(x1)) → R(x1)
R(s(x1)) → R(x1)
R(n(x1)) → R(x1)

The TRS R consists of the following rules:

r(r(x1)) → s(r(x1))
r(s(x1)) → s(r(x1))
r(n(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
n(u(x1)) → u(n(x1))
t(r(u(x1))) → t(c(r(x1)))
t(s(u(x1))) → t(c(r(x1)))
t(n(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(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.

R(u(x1)) → R(x1)
R(s(x1)) → R(x1)
R(n(x1)) → R(x1)

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

POL(n(x₁)) = 1/4 + (4)x_1
POL(u(x₁)) = 2 + (4)x_1
POL(s(x₁)) = 1/4 + x_1
POL(R(x₁)) = (2)x_1

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

r(r(x1)) → s(r(x1))
r(s(x1)) → s(r(x1))
r(n(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
n(u(x1)) → u(n(x1))
t(r(u(x1))) → t(c(r(x1)))
t(s(u(x1))) → t(c(r(x1)))
t(n(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(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

            ↳ QDPOrderProof

          ↳ QDP

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

C(n(x1)) → C(x1)
C(u(x1)) → C(x1)
C(s(x1)) → C(x1)
C(r(x1)) → C(x1)

The TRS R consists of the following rules:

r(r(x1)) → s(r(x1))
r(s(x1)) → s(r(x1))
r(n(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
n(u(x1)) → u(n(x1))
t(r(u(x1))) → t(c(r(x1)))
t(s(u(x1))) → t(c(r(x1)))
t(n(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(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)

The remaining pairs can at least be oriented weakly.

C(u(x1)) → C(x1)
C(s(x1)) → C(x1)
C(r(x1)) → C(x1)

Used ordering: Polynomial interpretation [25,35]:

POL(C(x₁)) = (2)x_1
POL(n(x₁)) = 4 + (2)x_1
POL(u(x₁)) = (2)x_1
POL(s(x₁)) = x_1
POL(r(x₁)) = (2)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

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

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

C(u(x1)) → C(x1)
C(s(x1)) → C(x1)
C(r(x1)) → C(x1)

The TRS R consists of the following rules:

r(r(x1)) → s(r(x1))
r(s(x1)) → s(r(x1))
r(n(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
n(u(x1)) → u(n(x1))
t(r(u(x1))) → t(c(r(x1)))
t(s(u(x1))) → t(c(r(x1)))
t(n(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(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(u(x1)) → C(x1)
C(s(x1)) → C(x1)
C(r(x1)) → C(x1)

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

POL(C(x₁)) = (1/4)x_1
POL(u(x₁)) = 1 + (4)x_1
POL(s(x₁)) = 2 + (4)x_1
POL(r(x₁)) = 1/2 + (2)x_1

The value of delta used in the strict ordering is 1/8.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

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

r(r(x1)) → s(r(x1))
r(s(x1)) → s(r(x1))
r(n(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
n(u(x1)) → u(n(x1))
t(r(u(x1))) → t(c(r(x1)))
t(s(u(x1))) → t(c(r(x1)))
t(n(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(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

          ↳ QDP

            ↳ QDPOrderProof

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

T(r(u(x1))) → T(c(r(x1)))
T(s(u(x1))) → T(c(r(x1)))
T(n(u(x1))) → T(c(r(x1)))

The TRS R consists of the following rules:

r(r(x1)) → s(r(x1))
r(s(x1)) → s(r(x1))
r(n(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
n(u(x1)) → u(n(x1))
t(r(u(x1))) → t(c(r(x1)))
t(s(u(x1))) → t(c(r(x1)))
t(n(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(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.

T(n(u(x1))) → T(c(r(x1)))

The remaining pairs can at least be oriented weakly.

T(r(u(x1))) → T(c(r(x1)))
T(s(u(x1))) → T(c(r(x1)))

Used ordering: Polynomial interpretation [25,35]:

POL(T(x₁)) = (1/4)x_1
POL(c(x₁)) = x_1
POL(n(x₁)) = 1/4
POL(u(x₁)) = 0
POL(r(x₁)) = 0
POL(s(x₁)) = 0
POL(b(x₁)) = 0

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:

n(u(x1)) → u(n(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(n(x1)) → n(x1)
r(s(x1)) → s(r(x1))
r(r(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(n(x1)) → s(r(x1))
s(u(x1)) → u(s(x1))
r(u(x1)) → u(r(x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

T(r(u(x1))) → T(c(r(x1)))
T(s(u(x1))) → T(c(r(x1)))

The TRS R consists of the following rules:

r(r(x1)) → s(r(x1))
r(s(x1)) → s(r(x1))
r(n(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
n(u(x1)) → u(n(x1))
t(r(u(x1))) → t(c(r(x1)))
t(s(u(x1))) → t(c(r(x1)))
t(n(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(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.

T(r(u(x1))) → T(c(r(x1)))

The remaining pairs can at least be oriented weakly.

T(s(u(x1))) → T(c(r(x1)))

Used ordering: Polynomial interpretation [25,35]:

POL(T(x₁)) = (1/4)x_1
POL(c(x₁)) = x_1
POL(n(x₁)) = (4)x_1
POL(u(x₁)) = 1/4 + (4)x_1
POL(r(x₁)) = 1/4 + (4)x_1
POL(s(x₁)) = x_1
POL(b(x₁)) = 0

The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

n(u(x1)) → u(n(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(n(x1)) → n(x1)
r(s(x1)) → s(r(x1))
r(r(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(n(x1)) → s(r(x1))
s(u(x1)) → u(s(x1))
r(u(x1)) → u(r(x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

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

T(s(u(x1))) → T(c(r(x1)))

The TRS R consists of the following rules:

r(r(x1)) → s(r(x1))
r(s(x1)) → s(r(x1))
r(n(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
n(u(x1)) → u(n(x1))
t(r(u(x1))) → t(c(r(x1)))
t(s(u(x1))) → t(c(r(x1)))
t(n(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(x1)

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