Termination w.r.t. Q proof of /tmp/tmpUo4AYj/dup15.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(d(d(x1)))) → d(d(b(b(x1))))
a(a(x1)) → b(b(b(b(b(b(x1))))))
b(b(d(d(b(b(x1)))))) → a(a(c(c(x1))))
c(c(x1)) → d(d(x1))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

a(a(d(d(x1)))) → d(d(b(b(x1))))
a(a(x1)) → b(b(b(b(b(b(x1))))))
b(b(d(d(b(b(x1)))))) → a(a(c(c(x1))))
c(c(x1)) → d(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 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

B(b(d(d(b(b(x1)))))) → A(c(c(x1)))

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

POL(c(x₁)) = 1/4 + (13/4)x_1
POL(B(x₁)) = 4 + (1/2)x_1
POL(a(x₁)) = x_1
POL(A(x₁)) = 4 + (1/2)x_1
POL(d(x₁)) = 1/4 + (13/4)x_1
POL(b(x₁)) = x_1

The value of delta used in the strict ordering is 17/32.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

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

The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

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

POL(c(x₁)) = 4 + (4)x_1
POL(B(x₁)) = 4 + (1/4)x_1
POL(a(x₁)) = 4 + x_1
POL(A(x₁)) = 15/4 + (1/4)x_1
POL(b(x₁)) = 1/4 + x_1
POL(d(x₁)) = 4 + (4)x_1

The value of delta used in the strict ordering is 7/16.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPOrderProof

                        ↳ QDP

                          ↳ PisEmptyProof

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

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