Termination w.r.t. Q proof of /tmp/tmpWEfkq0/z110.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(x1)) → b(x1)
b(a(x1)) → a(b(x1))
b(b(c(x1))) → c(a(x1))
b(b(x1)) → a(a(a(x1)))
c(a(x1)) → b(a(c(x1)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

B(b(c(x1))) → A(x1)

Used ordering: POLO with Polynomial interpretation [25]:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

B(a(x1)) → A(b(x1))
A(a(x1)) → B(x1)

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

B(a(x1)) → A(b(x1))

Used ordering: Polynomial interpretation [25,35]:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

B(a(x1)) → A(b(x1))

The TRS R consists of the following rules:

a(a(x1)) → b(x1)
b(a(x1)) → a(b(x1))
b(b(c(x1))) → c(a(x1))
b(b(x1)) → a(a(a(x1)))
c(a(x1)) → b(a(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 0 SCCs with 1 less node.
We have reversed the following QTRS:
The set of rules R is

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

The set Q is empty.
We have obtained the following QTRS:

a(a(x)) → b(x)
a(b(x)) → b(a(x))
c(b(b(x))) → a(c(x))
b(b(x)) → a(a(a(x)))
a(c(x)) → c(a(b(x)))

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ QTRS Reverse

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

a(a(x)) → b(x)
a(b(x)) → b(a(x))
c(b(b(x))) → a(c(x))
b(b(x)) → a(a(a(x)))
a(c(x)) → c(a(b(x)))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

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

The set Q is empty.
We have obtained the following QTRS:

a(a(x)) → b(x)
a(b(x)) → b(a(x))
c(b(b(x))) → a(c(x))
b(b(x)) → a(a(a(x)))
a(c(x)) → c(a(b(x)))

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

    ↳ QTRS

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

a(a(x)) → b(x)
a(b(x)) → b(a(x))
c(b(b(x))) → a(c(x))
b(b(x)) → a(a(a(x)))
a(c(x)) → c(a(b(x)))

Q is empty.