Termination w.r.t. Q proof of /tmp/tmpjIeTyj/04.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(x1))
b(a(a(x1))) → a(a(a(b(x1))))
a(c(x1)) → c(b(x1))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

A(c(x1)) → B(x1)

Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x₁)) = 2·x₁
POL(B(x₁)) = 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(a(a(x1))) → A(a(a(b(x1))))
B(a(a(x1))) → A(b(x1))
A(a(b(x1))) → A(x1)
B(a(a(x1))) → A(a(b(x1)))
B(a(a(x1))) → B(x1)
A(a(b(x1))) → B(a(x1))

The TRS R consists of the following rules:

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

A(a(b(x1))) → A(x1)

The remaining pairs can at least be oriented weakly.

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

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

POL( A(x₁) ) = x₁

POL( c(x₁) ) = max{0, -1}

POL( b(x₁) ) = x₁ + 1

POL( B(x₁) ) = x₁ + 1

POL( a(x₁) ) = x₁

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

POL(A(x₁)) = 3 + 2·x₁
POL(B(x₁)) = 4·x₁
POL(a(x₁)) = 1 + x₁
POL(b(x₁)) = 2·x₁
POL(c(x₁)) = 0

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ PisEmptyProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

a(a(b(x1))) → b(a(x1))
b(a(a(x1))) → a(a(a(b(x1))))
a(c(x1)) → c(b(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.
We have reversed the following QTRS:
The set of rules R is

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

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

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

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

Q is empty.

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

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

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

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

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

Q is empty.