Termination w.r.t. Q proof of /tmp/tmpVgUm99/z081.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:

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

b(c(a(x1))) → a(b(x1))
b(b(b(x1))) → c(a(c(x1)))
c(d(x1)) → d(c(x1))
c(d(b(x1))) → d(c(c(x1)))
d(c(x1)) → b(b(b(x1)))
c(b(x1)) → d(a(x1))
d(b(c(x1))) → a(a(x1))
d(a(x1)) → 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 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

D(c(x1)) → B(b(b(x1)))

Used ordering: Polynomial interpretation [25,35]:

POL(C(x₁)) = 1/4 + x_1
POL(c(x₁)) = 1 + (2)x_1
POL(D(x₁)) = 1 + (1/2)x_1
POL(B(x₁)) = 3/4 + (1/4)x_1
POL(a(x₁)) = x_1
POL(d(x₁)) = 3 + (4)x_1
POL(b(x₁)) = 1 + (2)x_1

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

D(c(x1)) → B(b(b(x1)))

The TRS R consists of the following rules:

b(c(a(x1))) → a(b(x1))
b(b(b(x1))) → c(a(c(x1)))
c(d(x1)) → d(c(x1))
c(d(b(x1))) → d(c(c(x1)))
d(c(x1)) → b(b(b(x1)))
c(b(x1)) → d(a(x1))
d(b(c(x1))) → a(a(x1))
d(a(x1)) → 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 0 SCCs with 1 less node.