Termination Proof using AProVE

Term Rewriting System R:
[f, y, ys]
app(app(filter, f), nil) -> nil
app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys)

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

APP(app(filter, f), app(app(cons, y), ys)) -> APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
APP(app(filter, f), app(app(cons, y), ys)) -> APP(app(filtersub, app(f, y)), f)
APP(app(filter, f), app(app(cons, y), ys)) -> APP(filtersub, app(f, y))
APP(app(filter, f), app(app(cons, y), ys)) -> APP(f, y)
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) -> APP(app(cons, y), app(app(filter, f), ys))
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) -> APP(filter, f)
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) -> APP(filter, f)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

APP(app(app(filtersub, false), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)
APP(app(filter, f), app(app(cons, y), ys)) -> APP(f, y)
APP(app(filter, f), app(app(cons, y), ys)) -> APP(app(filtersub, app(f, y)), f)
APP(app(filter, f), app(app(cons, y), ys)) -> APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))

Rules:

app(app(filter, f), nil) -> nil
app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys)

Strategy:

innermost

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

APP(app(filter, f), app(app(cons, y), ys)) -> APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))

four new Dependency Pairs are created:

APP(app(filter, app(filter, f'')), app(app(cons, nil), ys)) -> APP(app(app(filtersub, nil), app(filter, f'')), app(app(cons, nil), ys))
APP(app(filter, app(filter, f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(app(filtersub, app(f'', y'')), f''), app(app(cons, y''), ys''))), app(filter, f'')), app(app(cons, app(app(cons, y''), ys'')), ys))
APP(app(filter, app(app(filtersub, true), f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(cons, y''), app(app(filter, f''), ys''))), app(app(filtersub, true), f'')), app(app(cons, app(app(cons, y''), ys'')), ys))
APP(app(filter, app(app(filtersub, false), f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(filter, f''), ys'')), app(app(filtersub, false), f'')), app(app(cons, app(app(cons, y''), ys'')), ys))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

APP(app(filter, app(app(filtersub, false), f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(filter, f''), ys'')), app(app(filtersub, false), f'')), app(app(cons, app(app(cons, y''), ys'')), ys))
APP(app(filter, app(app(filtersub, true), f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(cons, y''), app(app(filter, f''), ys''))), app(app(filtersub, true), f'')), app(app(cons, app(app(cons, y''), ys'')), ys))
APP(app(filter, app(filter, f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(app(filtersub, app(f'', y'')), f''), app(app(cons, y''), ys''))), app(filter, f'')), app(app(cons, app(app(cons, y''), ys'')), ys))
APP(app(filter, app(filter, f'')), app(app(cons, nil), ys)) -> APP(app(app(filtersub, nil), app(filter, f'')), app(app(cons, nil), ys))
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)
APP(app(filter, f), app(app(cons, y), ys)) -> APP(f, y)
APP(app(filter, f), app(app(cons, y), ys)) -> APP(app(filtersub, app(f, y)), f)
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)

Rules:

app(app(filter, f), nil) -> nil
app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys)

Strategy:

innermost

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

APP(app(filter, f), app(app(cons, y), ys)) -> APP(app(filtersub, app(f, y)), f)

four new Dependency Pairs are created:

APP(app(filter, app(filter, f'')), app(app(cons, nil), ys)) -> APP(app(filtersub, nil), app(filter, f''))
APP(app(filter, app(filter, f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(filtersub, app(app(app(filtersub, app(f'', y'')), f''), app(app(cons, y''), ys''))), app(filter, f''))
APP(app(filter, app(app(filtersub, true), f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(filtersub, app(app(cons, y''), app(app(filter, f''), ys''))), app(app(filtersub, true), f''))
APP(app(filter, app(app(filtersub, false), f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(filtersub, app(app(filter, f''), ys'')), app(app(filtersub, false), f''))

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(filter, app(app(filtersub, true), f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(cons, y''), app(app(filter, f''), ys''))), app(app(filtersub, true), f'')), app(app(cons, app(app(cons, y''), ys'')), ys))
APP(app(filter, app(filter, f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(app(filtersub, app(f'', y'')), f''), app(app(cons, y''), ys''))), app(filter, f'')), app(app(cons, app(app(cons, y''), ys'')), ys))
APP(app(filter, app(filter, f'')), app(app(cons, nil), ys)) -> APP(app(app(filtersub, nil), app(filter, f'')), app(app(cons, nil), ys))
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)
APP(app(filter, f), app(app(cons, y), ys)) -> APP(f, y)
APP(app(filter, app(app(filtersub, false), f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(filter, f''), ys'')), app(app(filtersub, false), f'')), app(app(cons, app(app(cons, y''), ys'')), ys))

Rules:

app(app(filter, f), nil) -> nil
app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys)

Strategy:

innermost

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

APP(app(filter, app(filter, f'')), app(app(cons, nil), ys)) -> APP(app(app(filtersub, nil), app(filter, f'')), app(app(cons, nil), ys))

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(filter, app(app(filtersub, false), f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(filter, f''), ys'')), app(app(filtersub, false), f'')), app(app(cons, app(app(cons, y''), ys'')), ys))
APP(app(filter, app(filter, f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(app(filtersub, app(f'', y'')), f''), app(app(cons, y''), ys''))), app(filter, f'')), app(app(cons, app(app(cons, y''), ys'')), ys))
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)
APP(app(filter, f), app(app(cons, y), ys)) -> APP(f, y)
APP(app(filter, app(app(filtersub, true), f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(cons, y''), app(app(filter, f''), ys''))), app(app(filtersub, true), f'')), app(app(cons, app(app(cons, y''), ys'')), ys))

Rules:

app(app(filter, f), nil) -> nil
app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys)

Strategy:

innermost

The following dependency pair can be strictly oriented:

APP(app(filter, f), app(app(cons, y), ys)) -> APP(f, y)

Additionally, the following usable rules for innermost can be oriented:

app(app(filter, f), nil) -> nil
app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(filter) = 0

POL(cons) = 0

POL(false) = 0

POL(filtersub) = 0

POL(nil) = 0

POL(true) = 0

POL(app(x₁, x₂)) = 1 + x₂

POL(APP(x₁, x₂)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 5

                 ↳Polynomial Ordering

Dependency Pairs:

APP(app(filter, app(app(filtersub, false), f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(filter, f''), ys'')), app(app(filtersub, false), f'')), app(app(cons, app(app(cons, y''), ys'')), ys))
APP(app(filter, app(filter, f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(app(filtersub, app(f'', y'')), f''), app(app(cons, y''), ys''))), app(filter, f'')), app(app(cons, app(app(cons, y''), ys'')), ys))
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)
APP(app(filter, app(app(filtersub, true), f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(cons, y''), app(app(filter, f''), ys''))), app(app(filtersub, true), f'')), app(app(cons, app(app(cons, y''), ys'')), ys))

Rules:

app(app(filter, f), nil) -> nil
app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys)

Strategy:

innermost

The following dependency pairs can be strictly oriented:

APP(app(app(filtersub, false), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)

Additionally, the following usable rules for innermost can be oriented:

app(app(filter, f), nil) -> nil
app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(filter) = 0

POL(cons) = 0

POL(false) = 0

POL(filtersub) = 0

POL(nil) = 0

POL(true) = 0

POL(app(x₁, x₂)) = 1 + x₂

POL(APP(x₁, x₂)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 6

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

Rules:

app(app(filter, f), nil) -> nil
app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys)

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:02 minutes

POL(filter)	= 0
POL(cons)	= 0
POL(false)	= 0
POL(filtersub)	= 0
POL(nil)	= 0
POL(true)	= 0
POL(app(x₁, x₂))	= 1 + x₂
POL(APP(x₁, x₂))	= x₁