Termination w.r.t. Q proof of /tmp/tmpFbpgV0/dup17.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(a(a(x1)))) → b(b(x1))
b(b(a(a(x1)))) → a(a(b(b(x1))))
b(b(b(b(c(c(x1)))))) → c(c(a(a(x1))))
b(b(b(b(x1)))) → a(a(a(a(a(a(x1))))))
c(c(a(a(x1)))) → b(b(a(a(c(c(x1))))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

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

POL(c(x₁)) = (7/4)x_1
POL(B(x₁)) = 4 + (3/4)x_1
POL(a(x₁)) = 3/4 + x_1
POL(A(x₁)) = 7/2 + (3/4)x_1
POL(b(x₁)) = 5/4 + x_1

The value of delta used in the strict ordering is 1/8.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ PisEmptyProof

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

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