Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
app(app(:, app(app(:, x), y)), z) -> app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) -> app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) -> app(app(:, app(app(g, z), y)), app(app(+, x), a))

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

APP(app(:, app(app(:, x), y)), z) -> APP(app(:, x), app(app(:, y), z))
APP(app(:, app(app(:, x), y)), z) -> APP(app(:, y), z)
APP(app(:, app(app(:, x), y)), z) -> APP(:, y)
APP(app(:, app(app(+, x), y)), z) -> APP(app(+, app(app(:, x), z)), app(app(:, y), z))
APP(app(:, app(app(+, x), y)), z) -> APP(+, app(app(:, x), z))
APP(app(:, app(app(+, x), y)), z) -> APP(app(:, x), z)
APP(app(:, app(app(+, x), y)), z) -> APP(:, x)
APP(app(:, app(app(+, x), y)), z) -> APP(app(:, y), z)
APP(app(:, app(app(+, x), y)), z) -> APP(:, y)
APP(app(:, z), app(app(+, x), app(f, y))) -> APP(app(:, app(app(g, z), y)), app(app(+, x), a))
APP(app(:, z), app(app(+, x), app(f, y))) -> APP(:, app(app(g, z), y))
APP(app(:, z), app(app(+, x), app(f, y))) -> APP(app(g, z), y)
APP(app(:, z), app(app(+, x), app(f, y))) -> APP(g, z)
APP(app(:, z), app(app(+, x), app(f, y))) -> APP(app(+, x), a)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

APP(app(:, z), app(app(+, x), app(f, y))) -> APP(app(+, x), a)
APP(app(:, z), app(app(+, x), app(f, y))) -> APP(app(g, z), y)
APP(app(:, z), app(app(+, x), app(f, y))) -> APP(app(:, app(app(g, z), y)), app(app(+, x), a))
APP(app(:, app(app(+, x), y)), z) -> APP(app(:, y), z)
APP(app(:, app(app(+, x), y)), z) -> APP(app(:, x), z)
APP(app(:, app(app(+, x), y)), z) -> APP(app(+, app(app(:, x), z)), app(app(:, y), z))
APP(app(:, app(app(:, x), y)), z) -> APP(app(:, y), z)
APP(app(:, app(app(:, x), y)), z) -> APP(app(:, x), app(app(:, y), z))

Rules:

app(app(:, app(app(:, x), y)), z) -> app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) -> app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) -> app(app(:, app(app(g, z), y)), app(app(+, x), a))

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

APP(app(:, z), app(app(+, x), app(f, y))) -> APP(app(:, app(app(g, z), y)), app(app(+, x), a))

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

APP(app(:, z), app(app(+, x), app(f, y))) -> APP(app(g, z), y)
APP(app(:, app(app(+, x), y)), z) -> APP(app(:, y), z)
APP(app(:, app(app(+, x), y)), z) -> APP(app(:, x), z)
APP(app(:, app(app(+, x), y)), z) -> APP(app(+, app(app(:, x), z)), app(app(:, y), z))
APP(app(:, app(app(:, x), y)), z) -> APP(app(:, y), z)
APP(app(:, app(app(:, x), y)), z) -> APP(app(:, x), app(app(:, y), z))
APP(app(:, z), app(app(+, x), app(f, y))) -> APP(app(+, x), a)

Rules:

app(app(:, app(app(:, x), y)), z) -> app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) -> app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) -> app(app(:, app(app(g, z), y)), app(app(+, x), a))

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

APP(app(:, z), app(app(+, x), app(f, y))) -> APP(app(g, z), y)

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 3

                 ↳Narrowing Transformation

Dependency Pairs:

APP(app(:, z), app(app(+, x), app(f, y))) -> APP(app(+, x), a)
APP(app(:, app(app(+, x), y)), z) -> APP(app(:, x), z)
APP(app(:, app(app(+, x), y)), z) -> APP(app(+, app(app(:, x), z)), app(app(:, y), z))
APP(app(:, app(app(:, x), y)), z) -> APP(app(:, y), z)
APP(app(:, app(app(:, x), y)), z) -> APP(app(:, x), app(app(:, y), z))
APP(app(:, app(app(+, x), y)), z) -> APP(app(:, y), z)

Rules:

app(app(:, app(app(:, x), y)), z) -> app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) -> app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) -> app(app(:, app(app(g, z), y)), app(app(+, x), a))

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

APP(app(:, z), app(app(+, x), app(f, y))) -> APP(app(+, x), a)

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 4

                 ↳Polynomial Ordering

Dependency Pairs:

APP(app(:, app(app(+, x), y)), z) -> APP(app(:, y), z)
APP(app(:, app(app(+, x), y)), z) -> APP(app(+, app(app(:, x), z)), app(app(:, y), z))
APP(app(:, app(app(:, x), y)), z) -> APP(app(:, y), z)
APP(app(:, app(app(:, x), y)), z) -> APP(app(:, x), app(app(:, y), z))
APP(app(:, app(app(+, x), y)), z) -> APP(app(:, x), z)

Rules:

app(app(:, app(app(:, x), y)), z) -> app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) -> app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) -> app(app(:, app(app(g, z), y)), app(app(+, x), a))

The following dependency pair can be strictly oriented:

APP(app(:, app(app(+, x), y)), z) -> APP(app(+, app(app(:, x), z)), app(app(:, y), z))

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

app(app(:, app(app(:, x), y)), z) -> app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) -> app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) -> app(app(:, app(app(g, z), y)), app(app(+, x), a))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(:) = 1

POL(g) = 0

POL(a) = 0

POL(app(x₁, x₂)) = x₁

POL(+) = 0

POL(f) = 0

POL(APP(x₁, x₂)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 5

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

APP(app(:, app(app(+, x), y)), z) -> APP(app(:, y), z)
APP(app(:, app(app(:, x), y)), z) -> APP(app(:, y), z)
APP(app(:, app(app(:, x), y)), z) -> APP(app(:, x), app(app(:, y), z))
APP(app(:, app(app(+, x), y)), z) -> APP(app(:, x), z)

Rules:

app(app(:, app(app(:, x), y)), z) -> app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) -> app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) -> app(app(:, app(app(g, z), y)), app(app(+, x), a))

Termination of R could not be shown.
Duration:
0:00 minutes

POL(:)	= 1
POL(g)	= 0
POL(a)	= 0
POL(app(x₁, x₂))	= x₁
POL(+)	= 0
POL(f)	= 0
POL(APP(x₁, x₂))	= x₁