Termination w.r.t. Q proof of /tmp/tmp2sU-PU/z119.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:

a(b(x1)) → b(d(x1))
a(c(x1)) → d(d(d(x1)))
b(d(x1)) → a(c(b(x1)))
c(f(x1)) → d(d(c(x1)))
d(d(x1)) → f(x1)
f(f(x1)) → a(x1)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a(b(x1)) → b(d(x1))
a(c(x1)) → d(d(d(x1)))
b(d(x1)) → a(c(b(x1)))
c(f(x1)) → d(d(c(x1)))
d(d(x1)) → f(x1)
f(f(x1)) → a(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:

A(c(x1)) → D(d(d(x1)))
F(f(x1)) → A(x1)
A(b(x1)) → B(d(x1))
C(f(x1)) → C(x1)
A(b(x1)) → D(x1)
B(d(x1)) → A(c(b(x1)))
A(c(x1)) → D(x1)
B(d(x1)) → C(b(x1))
C(f(x1)) → D(d(c(x1)))
A(c(x1)) → D(d(x1))
D(d(x1)) → F(x1)
B(d(x1)) → B(x1)
C(f(x1)) → D(c(x1))

The TRS R consists of the following rules:

a(b(x1)) → b(d(x1))
a(c(x1)) → d(d(d(x1)))
b(d(x1)) → a(c(b(x1)))
c(f(x1)) → d(d(c(x1)))
d(d(x1)) → f(x1)
f(f(x1)) → a(x1)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

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

A(c(x1)) → D(d(d(x1)))
F(f(x1)) → A(x1)
A(b(x1)) → B(d(x1))
C(f(x1)) → C(x1)
A(b(x1)) → D(x1)
B(d(x1)) → A(c(b(x1)))
A(c(x1)) → D(x1)
B(d(x1)) → C(b(x1))
C(f(x1)) → D(d(c(x1)))
A(c(x1)) → D(d(x1))
D(d(x1)) → F(x1)
B(d(x1)) → B(x1)
C(f(x1)) → D(c(x1))

The TRS R consists of the following rules:

a(b(x1)) → b(d(x1))
a(c(x1)) → d(d(d(x1)))
b(d(x1)) → a(c(b(x1)))
c(f(x1)) → d(d(c(x1)))
d(d(x1)) → f(x1)
f(f(x1)) → a(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.

A(b(x1)) → D(x1)

The remaining pairs can at least be oriented weakly.

A(c(x1)) → D(d(d(x1)))
F(f(x1)) → A(x1)
A(b(x1)) → B(d(x1))
C(f(x1)) → C(x1)
B(d(x1)) → A(c(b(x1)))
A(c(x1)) → D(x1)
B(d(x1)) → C(b(x1))
C(f(x1)) → D(d(c(x1)))
A(c(x1)) → D(d(x1))
D(d(x1)) → F(x1)
B(d(x1)) → B(x1)
C(f(x1)) → D(c(x1))

Used ordering: Polynomial interpretation [25,35]:

POL(C(x₁)) = (4)x_1
POL(c(x₁)) = x_1
POL(f(x₁)) = x_1
POL(D(x₁)) = (4)x_1
POL(B(x₁)) = 4 + (4)x_1
POL(a(x₁)) = x_1
POL(A(x₁)) = (4)x_1
POL(d(x₁)) = x_1
POL(b(x₁)) = 1 + x_1
POL(F(x₁)) = (4)x_1

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

c(f(x1)) → d(d(c(x1)))
a(c(x1)) → d(d(d(x1)))
d(d(x1)) → f(x1)
b(d(x1)) → a(c(b(x1)))
a(b(x1)) → b(d(x1))
f(f(x1)) → a(x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

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

B(d(x1)) → C(b(x1))
C(f(x1)) → D(d(c(x1)))
A(c(x1)) → D(d(x1))
F(f(x1)) → A(x1)
A(c(x1)) → D(d(d(x1)))
A(b(x1)) → B(d(x1))
D(d(x1)) → F(x1)
C(f(x1)) → C(x1)
B(d(x1)) → B(x1)
B(d(x1)) → A(c(b(x1)))
A(c(x1)) → D(x1)
C(f(x1)) → D(c(x1))

The TRS R consists of the following rules:

a(b(x1)) → b(d(x1))
a(c(x1)) → d(d(d(x1)))
b(d(x1)) → a(c(b(x1)))
c(f(x1)) → d(d(c(x1)))
d(d(x1)) → f(x1)
f(f(x1)) → a(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.

B(d(x1)) → C(b(x1))
A(c(x1)) → D(d(x1))
A(c(x1)) → D(d(d(x1)))
C(f(x1)) → C(x1)
B(d(x1)) → B(x1)
A(c(x1)) → D(x1)
C(f(x1)) → D(c(x1))

The remaining pairs can at least be oriented weakly.

C(f(x1)) → D(d(c(x1)))
F(f(x1)) → A(x1)
A(b(x1)) → B(d(x1))
D(d(x1)) → F(x1)
B(d(x1)) → A(c(b(x1)))

Used ordering: Polynomial interpretation [25,35]:

POL(C(x₁)) = x_1
POL(f(x₁)) = 1/2 + x_1
POL(c(x₁)) = x_1
POL(B(x₁)) = (4)x_1
POL(D(x₁)) = 1/4 + x_1
POL(a(x₁)) = 1 + x_1
POL(A(x₁)) = 1 + x_1
POL(d(x₁)) = 1/4 + x_1
POL(b(x₁)) = (4)x_1
POL(F(x₁)) = 1/2 + x_1

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

c(f(x1)) → d(d(c(x1)))
a(c(x1)) → d(d(d(x1)))
d(d(x1)) → f(x1)
b(d(x1)) → a(c(b(x1)))
a(b(x1)) → b(d(x1))
f(f(x1)) → a(x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

C(f(x1)) → D(d(c(x1)))
F(f(x1)) → A(x1)
A(b(x1)) → B(d(x1))
D(d(x1)) → F(x1)
B(d(x1)) → A(c(b(x1)))

The TRS R consists of the following rules:

a(b(x1)) → b(d(x1))
a(c(x1)) → d(d(d(x1)))
b(d(x1)) → a(c(b(x1)))
c(f(x1)) → d(d(c(x1)))
d(d(x1)) → f(x1)
f(f(x1)) → a(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

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

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

A(b(x1)) → B(d(x1))
B(d(x1)) → A(c(b(x1)))

The TRS R consists of the following rules:

a(b(x1)) → b(d(x1))
a(c(x1)) → d(d(d(x1)))
b(d(x1)) → a(c(b(x1)))
c(f(x1)) → d(d(c(x1)))
d(d(x1)) → f(x1)
f(f(x1)) → a(x1)

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