Termination w.r.t. Q proof of /tmp/tmpRrgeBh/beans4.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(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
b(c(a(x1))) → a(b(c(x1)))
c(b(x1)) → b(a(x1))
a(c(b(x1))) → c(b(a(x1)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

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

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

POL(C(x₁)) = 7/4 + (1/2)x_1
POL(c(x₁)) = 1/4 + (1/4)x_1
POL(B(x₁)) = 7/4
POL(a(x₁)) = x_1
POL(A(x₁)) = 5/4 + (2)x_1
POL(b(x₁)) = 1/4

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

POL(C(x₁)) = (4)x_1
POL(c(x₁)) = 4 + (4)x_1
POL(a(x₁)) = 4 + (4)x_1
POL(A(x₁)) = (4)x_1
POL(b(x₁)) = x_1

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                  ↳ QDP

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

A(c(b(x1))) → C(b(a(x1)))

Used ordering: Polynomial interpretation [25,35]:

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

The value of delta used in the strict ordering is 5.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                  ↳ QDP

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

A(c(b(x1))) → C(b(a(x1)))

The TRS R consists of the following rules:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

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

B(a(a(x1))) → B(c(x1))
B(c(a(x1))) → B(c(x1))

The TRS R consists of the following rules:

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

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

POL(c(x₁)) = 3 + (4)x_1
POL(B(x₁)) = (4)x_1
POL(a(x₁)) = 4 + (4)x_1
POL(b(x₁)) = 3 + (2)x_1

The value of delta used in the strict ordering is 64.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

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