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.



   R
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.


   R
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:



   R
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:



   R
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:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 4
Argument Filtering and 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)


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: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
APP(x1, x2) -> x1
app(x1, x2) -> app(x2)


   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 5
Argument Filtering and 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)


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: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
APP(x1, x2) -> APP(x1, x2)
app(x1, x2) -> app(x2)


   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 6
Remaining Obligation(s)




The following remains to be proven:
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(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



Innermost Termination of R could not be shown.
Duration:
0:03 minutes