Termination w.r.t. Q proof of /tmp/tmpJ5xW8o/z122.srs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

POL(C(x₁)) = 4 + (4)x_1
POL(c(x₁)) = 11/4 + x_1
POL(f(x₁)) = 9/4 + x_1
POL(D(x₁)) = (4)x_1
POL(B(x₁)) = 2 + (4)x_1
POL(a(x₁)) = 5/2 + x_1
POL(A(x₁)) = 3 + (4)x_1
POL(d(x₁)) = 7/4 + x_1
POL(b(x₁)) = 9/4 + x_1
POL(F(x₁)) = 1/4 + (4)x_1

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

a(a(x1)) → b(c(x1))
b(b(x1)) → c(d(x1))
c(c(x1)) → d(d(d(x1)))
d(c(x1)) → b(f(x1))
d(d(d(x1))) → a(c(x1))
f(f(x1)) → f(b(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 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

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

F(f(x1)) → F(b(x1))

The TRS R consists of the following rules:

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

F(f(x1)) → F(b(x1))

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

POL(f(x₁)) = 1/4 + (4)x_1
POL(c(x₁)) = (3/4)x_1
POL(a(x₁)) = 0
POL(b(x₁)) = 0
POL(d(x₁)) = 0
POL(F(x₁)) = (4)x_1

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ PisEmptyProof

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

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