Termination w.r.t. Q proof of /tmp/tmp2c5vLi/4.03.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:

+(x, 0) → x
+(minus(x), x) → 0
minus(0) → 0
minus(minus(x)) → x
minus(+(x, y)) → +(minus(y), minus(x))
*(x, 1) → x
*(x, 0) → 0
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(x, minus(y)) → minus(*(x, y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

+(x, 0) → x
+(minus(x), x) → 0
minus(0) → 0
minus(minus(x)) → x
minus(+(x, y)) → +(minus(y), minus(x))
*(x, 1) → x
*(x, 0) → 0
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(x, minus(y)) → minus(*(x, y))

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:

*¹(x, +(y, z)) → *¹(x, y)
MINUS(+(x, y)) → +¹(minus(y), minus(x))
*¹(x, minus(y)) → MINUS(*(x, y))
*¹(x, +(y, z)) → *¹(x, z)
*¹(x, minus(y)) → *¹(x, y)
MINUS(+(x, y)) → MINUS(y)
MINUS(+(x, y)) → MINUS(x)
*¹(x, +(y, z)) → +¹(*(x, y), *(x, z))

The TRS R consists of the following rules:

+(x, 0) → x
+(minus(x), x) → 0
minus(0) → 0
minus(minus(x)) → x
minus(+(x, y)) → +(minus(y), minus(x))
*(x, 1) → x
*(x, 0) → 0
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(x, minus(y)) → minus(*(x, y))

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:

*¹(x, +(y, z)) → *¹(x, y)
MINUS(+(x, y)) → +¹(minus(y), minus(x))
*¹(x, minus(y)) → MINUS(*(x, y))
*¹(x, +(y, z)) → *¹(x, z)
*¹(x, minus(y)) → *¹(x, y)
MINUS(+(x, y)) → MINUS(y)
MINUS(+(x, y)) → MINUS(x)
*¹(x, +(y, z)) → +¹(*(x, y), *(x, z))

The TRS R consists of the following rules:

+(x, 0) → x
+(minus(x), x) → 0
minus(0) → 0
minus(minus(x)) → x
minus(+(x, y)) → +(minus(y), minus(x))
*(x, 1) → x
*(x, 0) → 0
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(x, minus(y)) → minus(*(x, y))

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(y)
MINUS(+(x, y)) → MINUS(x)

The TRS R consists of the following rules:

+(x, 0) → x
+(minus(x), x) → 0
minus(0) → 0
minus(minus(x)) → x
minus(+(x, y)) → +(minus(y), minus(x))
*(x, 1) → x
*(x, 0) → 0
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(x, minus(y)) → minus(*(x, y))

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

                ↳ QDPSizeChangeProof

          ↳ QDP

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

MINUS(+(x, y)) → MINUS(y)
MINUS(+(x, y)) → MINUS(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:

MINUS(+(x, y)) → MINUS(y)
The graph contains the following edges 1 > 1
MINUS(+(x, y)) → MINUS(x)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

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

*¹(x, +(y, z)) → *¹(x, y)
*¹(x, +(y, z)) → *¹(x, z)
*¹(x, minus(y)) → *¹(x, y)

The TRS R consists of the following rules:

+(x, 0) → x
+(minus(x), x) → 0
minus(0) → 0
minus(minus(x)) → x
minus(+(x, y)) → +(minus(y), minus(x))
*(x, 1) → x
*(x, 0) → 0
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(x, minus(y)) → minus(*(x, y))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

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

*¹(x, +(y, z)) → *¹(x, y)
*¹(x, +(y, z)) → *¹(x, z)
*¹(x, minus(y)) → *¹(x, y)

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

*¹(x, +(y, z)) → *¹(x, y)
The graph contains the following edges 1 >= 1, 2 > 2
*¹(x, +(y, z)) → *¹(x, z)
The graph contains the following edges 1 >= 1, 2 > 2
*¹(x, minus(y)) → *¹(x, y)
The graph contains the following edges 1 >= 1, 2 > 2