Termination Proof using AProVE

Term Rewriting System R:
[x, y]
app(D, t) -> 1
app(D, constant) -> 0
app(D, app(app(+, x), y)) -> app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) -> app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) -> app(app(-, app(D, x)), app(D, y))

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

APP(D, app(app(+, x), y)) -> APP(app(+, app(D, x)), app(D, y))
APP(D, app(app(+, x), y)) -> APP(+, app(D, x))
APP(D, app(app(+, x), y)) -> APP(D, x)
APP(D, app(app(+, x), y)) -> APP(D, y)
APP(D, app(app(*, x), y)) -> APP(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
APP(D, app(app(*, x), y)) -> APP(+, app(app(*, y), app(D, x)))
APP(D, app(app(*, x), y)) -> APP(app(*, y), app(D, x))
APP(D, app(app(*, x), y)) -> APP(*, y)
APP(D, app(app(*, x), y)) -> APP(D, x)
APP(D, app(app(*, x), y)) -> APP(app(*, x), app(D, y))
APP(D, app(app(*, x), y)) -> APP(D, y)
APP(D, app(app(-, x), y)) -> APP(app(-, app(D, x)), app(D, y))
APP(D, app(app(-, x), y)) -> APP(-, app(D, x))
APP(D, app(app(-, x), y)) -> APP(D, x)
APP(D, app(app(-, x), y)) -> APP(D, y)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

Dependency Pairs:

APP(D, app(app(-, x), y)) -> APP(D, y)
APP(D, app(app(-, x), y)) -> APP(D, x)
APP(D, app(app(-, x), y)) -> APP(app(-, app(D, x)), app(D, y))
APP(D, app(app(*, x), y)) -> APP(D, y)
APP(D, app(app(*, x), y)) -> APP(app(*, x), app(D, y))
APP(D, app(app(*, x), y)) -> APP(D, x)
APP(D, app(app(*, x), y)) -> APP(app(*, y), app(D, x))
APP(D, app(app(*, x), y)) -> APP(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
APP(D, app(app(+, x), y)) -> APP(D, y)
APP(D, app(app(+, x), y)) -> APP(D, x)
APP(D, app(app(+, x), y)) -> APP(app(+, app(D, x)), app(D, y))

Rules:

app(D, t) -> 1
app(D, constant) -> 0
app(D, app(app(+, x), y)) -> app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) -> app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) -> app(app(-, app(D, x)), app(D, y))

The following dependency pairs can be strictly oriented:

APP(D, app(app(-, x), y)) -> APP(app(-, app(D, x)), app(D, y))
APP(D, app(app(*, x), y)) -> APP(app(*, x), app(D, y))
APP(D, app(app(*, x), y)) -> APP(app(*, y), app(D, x))
APP(D, app(app(*, x), y)) -> APP(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
APP(D, app(app(+, x), y)) -> APP(app(+, app(D, x)), app(D, y))

Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

app(D, t) -> 1
app(D, constant) -> 0
app(D, app(app(+, x), y)) -> app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) -> app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) -> app(app(-, app(D, x)), app(D, y))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(t) = 0

POL(0) = 0

POL(1) = 0

POL(constant) = 0

POL(*) = 0

POL(D) = 1

POL(-) = 0

POL(app(x₁, x₂)) = 0

POL(+) = 0

POL(APP(x₁, x₂)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳Polynomial Ordering

Dependency Pairs:

APP(D, app(app(-, x), y)) -> APP(D, y)
APP(D, app(app(-, x), y)) -> APP(D, x)
APP(D, app(app(*, x), y)) -> APP(D, y)
APP(D, app(app(*, x), y)) -> APP(D, x)
APP(D, app(app(+, x), y)) -> APP(D, y)
APP(D, app(app(+, x), y)) -> APP(D, x)

Rules:

app(D, t) -> 1
app(D, constant) -> 0
app(D, app(app(+, x), y)) -> app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) -> app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) -> app(app(-, app(D, x)), app(D, y))

The following dependency pairs can be strictly oriented:

APP(D, app(app(+, x), y)) -> APP(D, y)
APP(D, app(app(+, x), y)) -> APP(D, x)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(*) = 0

POL(D) = 0

POL(-) = 0

POL(APP(x₁, x₂)) = 1 + x₂

POL(app(x₁, x₂)) = x₁ + x₂

POL(+) = 1

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳Polo

...

               →DP Problem 3

                 ↳Polynomial Ordering

Dependency Pairs:

APP(D, app(app(-, x), y)) -> APP(D, y)
APP(D, app(app(-, x), y)) -> APP(D, x)
APP(D, app(app(*, x), y)) -> APP(D, y)
APP(D, app(app(*, x), y)) -> APP(D, x)

Rules:

app(D, t) -> 1
app(D, constant) -> 0
app(D, app(app(+, x), y)) -> app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) -> app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) -> app(app(-, app(D, x)), app(D, y))

The following dependency pairs can be strictly oriented:

APP(D, app(app(-, x), y)) -> APP(D, y)
APP(D, app(app(-, x), y)) -> APP(D, x)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(*) = 0

POL(D) = 0

POL(-) = 1

POL(APP(x₁, x₂)) = 1 + x₂

POL(app(x₁, x₂)) = x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳Polo

...

               →DP Problem 4

                 ↳Polynomial Ordering

Dependency Pairs:

APP(D, app(app(*, x), y)) -> APP(D, y)
APP(D, app(app(*, x), y)) -> APP(D, x)

Rules:

app(D, t) -> 1
app(D, constant) -> 0
app(D, app(app(+, x), y)) -> app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) -> app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) -> app(app(-, app(D, x)), app(D, y))

The following dependency pairs can be strictly oriented:

APP(D, app(app(*, x), y)) -> APP(D, y)
APP(D, app(app(*, x), y)) -> APP(D, x)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(*) = 1

POL(D) = 0

POL(APP(x₁, x₂)) = 1 + x₂

POL(app(x₁, x₂)) = x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳Polo

...

               →DP Problem 5

                 ↳Dependency Graph

Dependency Pair:

Rules:

app(D, t) -> 1
app(D, constant) -> 0
app(D, app(app(+, x), y)) -> app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) -> app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) -> app(app(-, app(D, x)), app(D, y))

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes

POL(t)	= 0
POL(0)	= 0
POL(1)	= 0
POL(constant)	= 0
POL(*)	= 0
POL(D)	= 1
POL(-)	= 0
POL(app(x₁, x₂))	= 0
POL(+)	= 0
POL(APP(x₁, x₂))	= 1 + x₁