Termination w.r.t. Q proof of /tmp/tmp9SvbSb/z083.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:

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

b(d(b(x1))) → c(d(b(x1)))
b(a(c(x1))) → b(c(x1))
a(d(x1)) → d(c(x1))
b(b(b(x1))) → a(b(c(x1)))
d(c(x1)) → b(d(x1))
d(c(x1)) → d(b(d(x1)))
d(a(c(x1))) → b(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 7 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

D(c(x1)) → D(b(d(x1)))
D(c(x1)) → D(x1)

The TRS R consists of the following rules:

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

D(c(x1)) → D(x1)

The remaining pairs can at least be oriented weakly.

D(c(x1)) → D(b(d(x1)))

Used ordering: Polynomial interpretation [25,35]:

POL(c(x₁)) = 4 + (4)x_1
POL(D(x₁)) = (4)x_1
POL(a(x₁)) = 4 + (4)x_1
POL(b(x₁)) = 4 + (4)x_1
POL(d(x₁)) = x_1

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

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

D(c(x1)) → D(b(d(x1)))

The TRS R consists of the following rules:

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

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