Termination w.r.t. Q proof of /tmp/tmpsU9fL9/t012.trs

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:

minus(minus(x)) → x
minus(+(x, y)) → *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) → +(minus(minus(minus(x))), minus(minus(minus(y))))
f(minus(x)) → minus(minus(minus(f(x))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

minus(minus(x)) → x
minus(+(x, y)) → *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) → +(minus(minus(minus(x))), minus(minus(minus(y))))
f(minus(x)) → minus(minus(minus(f(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:

MINUS(*(x, y)) → MINUS(minus(minus(x)))
F(minus(x)) → F(x)
MINUS(+(x, y)) → MINUS(minus(y))
MINUS(*(x, y)) → MINUS(minus(minus(y)))
MINUS(*(x, y)) → MINUS(y)
MINUS(+(x, y)) → MINUS(minus(minus(y)))
MINUS(+(x, y)) → MINUS(minus(x))
F(minus(x)) → MINUS(f(x))
F(minus(x)) → MINUS(minus(f(x)))
MINUS(+(x, y)) → MINUS(y)
MINUS(*(x, y)) → MINUS(minus(y))
MINUS(+(x, y)) → MINUS(minus(minus(x)))
F(minus(x)) → MINUS(minus(minus(f(x))))
MINUS(+(x, y)) → MINUS(x)
MINUS(*(x, y)) → MINUS(minus(x))
MINUS(*(x, y)) → MINUS(x)

The TRS R consists of the following rules:

minus(minus(x)) → x
minus(+(x, y)) → *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) → +(minus(minus(minus(x))), minus(minus(minus(y))))
f(minus(x)) → minus(minus(minus(f(x))))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

MINUS(*(x, y)) → MINUS(minus(minus(x)))
F(minus(x)) → F(x)
MINUS(+(x, y)) → MINUS(minus(y))
MINUS(*(x, y)) → MINUS(minus(minus(y)))
MINUS(*(x, y)) → MINUS(y)
MINUS(+(x, y)) → MINUS(minus(minus(y)))
MINUS(+(x, y)) → MINUS(minus(x))
F(minus(x)) → MINUS(f(x))
F(minus(x)) → MINUS(minus(f(x)))
MINUS(+(x, y)) → MINUS(y)
MINUS(*(x, y)) → MINUS(minus(y))
MINUS(+(x, y)) → MINUS(minus(minus(x)))
F(minus(x)) → MINUS(minus(minus(f(x))))
MINUS(+(x, y)) → MINUS(x)
MINUS(*(x, y)) → MINUS(minus(x))
MINUS(*(x, y)) → MINUS(x)

The TRS R consists of the following rules:

minus(minus(x)) → x
minus(+(x, y)) → *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) → +(minus(minus(minus(x))), minus(minus(minus(y))))
f(minus(x)) → minus(minus(minus(f(x))))

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 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

MINUS(*(x, y)) → MINUS(minus(minus(x)))
MINUS(+(x, y)) → MINUS(minus(x))
MINUS(+(x, y)) → MINUS(minus(y))
MINUS(*(x, y)) → MINUS(y)
MINUS(*(x, y)) → MINUS(minus(minus(y)))
MINUS(+(x, y)) → MINUS(minus(minus(y)))
MINUS(+(x, y)) → MINUS(y)
MINUS(*(x, y)) → MINUS(minus(y))
MINUS(+(x, y)) → MINUS(minus(minus(x)))
MINUS(*(x, y)) → MINUS(minus(x))
MINUS(+(x, y)) → MINUS(x)
MINUS(*(x, y)) → MINUS(x)

The TRS R consists of the following rules:

minus(minus(x)) → x
minus(+(x, y)) → *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) → +(minus(minus(minus(x))), minus(minus(minus(y))))
f(minus(x)) → minus(minus(minus(f(x))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

          ↳ QDP

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

MINUS(+(x, y)) → MINUS(minus(x))
MINUS(*(x, y)) → MINUS(minus(minus(x)))
MINUS(+(x, y)) → MINUS(minus(y))
MINUS(*(x, y)) → MINUS(minus(minus(y)))
MINUS(*(x, y)) → MINUS(y)
MINUS(+(x, y)) → MINUS(minus(minus(y)))
MINUS(+(x, y)) → MINUS(y)
MINUS(*(x, y)) → MINUS(minus(y))
MINUS(+(x, y)) → MINUS(minus(minus(x)))
MINUS(+(x, y)) → MINUS(x)
MINUS(*(x, y)) → MINUS(minus(x))
MINUS(*(x, y)) → MINUS(x)

The TRS R consists of the following rules:

minus(minus(x)) → x
minus(+(x, y)) → *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) → +(minus(minus(minus(x))), minus(minus(minus(y))))

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:

MINUS(+(x, y)) → MINUS(minus(x))
MINUS(*(x, y)) → MINUS(minus(minus(x)))
MINUS(+(x, y)) → MINUS(minus(y))
MINUS(*(x, y)) → MINUS(minus(minus(y)))
MINUS(*(x, y)) → MINUS(y)
MINUS(+(x, y)) → MINUS(minus(minus(y)))
MINUS(+(x, y)) → MINUS(y)
MINUS(*(x, y)) → MINUS(minus(y))
MINUS(+(x, y)) → MINUS(minus(minus(x)))
MINUS(+(x, y)) → MINUS(x)
MINUS(*(x, y)) → MINUS(minus(x))
MINUS(*(x, y)) → MINUS(x)

Used ordering: POLO with Polynomial interpretation [25]:

POL(*(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(+(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(MINUS(x₁)) = x₁
POL(minus(x₁)) = x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

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

minus(minus(x)) → x
minus(+(x, y)) → *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) → +(minus(minus(minus(x))), minus(minus(minus(y))))

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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

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

F(minus(x)) → F(x)

The TRS R consists of the following rules:

minus(minus(x)) → x
minus(+(x, y)) → *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) → +(minus(minus(minus(x))), minus(minus(minus(y))))
f(minus(x)) → minus(minus(minus(f(x))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

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

F(minus(x)) → F(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

F(minus(x)) → F(x)
The graph contains the following edges 1 > 1