Termination w.r.t. Q proof of /tmp/tmpzKYcxp/z121.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:

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

          ↳ DependencyGraphProof

            ↳ QDP

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

D(b(x1)) → D(a(b(x1)))

The TRS R consists of the following rules:

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

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