Termination w.r.t. Q proof of /tmp/tmppFjo9q/z022.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(b(x1))) → b(a(b(c(a(x1)))))
b(a(x1)) → a(b(b(x1)))
b(c(a(x1))) → c(a(b(x1)))

Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

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

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

The TRS R consists of the following rules:

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

Used ordering: POLO with Polynomial interpretation [25]:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial Order [21,25] with Interpretation:

POL( A(x₁) ) = x₁ + 1

POL( c(x₁) ) = max{0, x₁ - 1}

POL( b(x₁) ) = x₁

POL( B(x₁) ) = x₁ + 1

POL( a(x₁) ) = x₁ + 1

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

A(a(b(x1))) → B(a(b(c(a(x1)))))

Used ordering: Polynomial Order [21,25] with Interpretation:

POL( A(x₁) ) = 0

POL( c(x₁) ) = 1

POL( b(x₁) ) = x₁

POL( B(x₁) ) = x₁

POL( a(x₁) ) = max{0, -1}

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

A(a(b(x1))) → B(a(b(c(a(x1)))))

The TRS R consists of the following rules:

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