Termination w.r.t. Q proof of /tmp/tmpMZkN2h/10.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(d(x1)) → d(b(x1))
a(x1) → b(b(b(x1)))
d(x1) → x1
a(x1) → x1
b(d(b(x1))) → a(d(x1))
b(c(x1)) → c(d(d(x1)))
a(c(x1)) → b(b(c(d(x1))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

a(d(x1)) → d(b(x1))
a(x1) → b(b(b(x1)))
d(x1) → x1
a(x1) → x1
b(d(b(x1))) → a(d(x1))
b(c(x1)) → c(d(d(x1)))
a(c(x1)) → b(b(c(d(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 6 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

POL(c(x₁)) = 0
POL(B(x₁)) = (2)x_1
POL(a(x₁)) = x_1
POL(A(x₁)) = (2)x_1
POL(d(x₁)) = 4 + (4)x_1
POL(b(x₁)) = x_1

The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

A(x1) → B(b(b(x1)))

Used ordering: Polynomial interpretation [25,35]:

POL(c(x₁)) = 0
POL(B(x₁)) = x_1
POL(a(x₁)) = 4 + x_1
POL(A(x₁)) = 2 + x_1
POL(b(x₁)) = 1 + x_1
POL(d(x₁)) = (4)x_1

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

A(x1) → B(b(b(x1)))

The TRS R consists of the following rules:

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