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

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs))), app(app(filter, f), xs))
APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs)))
APP(app(filter, f), app(app(cons, x), xs)) -> APP(if, app(f, x))
APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(cons, x), app(app(filter, f), xs))
APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(filter, f), xs)

Furthermore, R contains one SCC.


   R
DPs
       →DP Problem 1
Narrowing Transformation


Dependency Pairs:

APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs)))
APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs))), app(app(filter, f), xs))


Rules:


app(app(app(if, true), x), y) -> x
app(app(app(if, false), x), y) -> y
app(app(filter, f), nil) -> nil
app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs))), app(app(filter, f), xs))


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, x), xs)) -> APP(app(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs))), app(app(filter, f), xs))
eight new Dependency Pairs are created:

APP(app(filter, app(app(if, true), x'')), app(app(cons, x0), xs)) -> APP(app(app(if, x''), app(app(cons, x0), app(app(filter, app(app(if, true), x'')), xs))), app(app(filter, app(app(if, true), x'')), xs))
APP(app(filter, app(app(if, false), x'')), app(app(cons, x0), xs)) -> APP(app(app(if, x0), app(app(cons, x0), app(app(filter, app(app(if, false), x'')), xs))), app(app(filter, app(app(if, false), x'')), xs))
APP(app(filter, app(filter, f'')), app(app(cons, nil), xs)) -> APP(app(app(if, nil), app(app(cons, nil), app(app(filter, app(filter, f'')), xs))), app(app(filter, app(filter, f'')), xs))
APP(app(filter, app(filter, f'')), app(app(cons, app(app(cons, x''), xs'')), xs)) -> APP(app(app(if, app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs''))), app(app(cons, app(app(cons, x''), xs'')), app(app(filter, app(filter, f'')), xs))), app(app(filter, app(filter, f'')), xs))
APP(app(filter, f''), app(app(cons, x), nil)) -> APP(app(app(if, app(f'', x)), app(app(cons, x), nil)), app(app(filter, f''), nil))
APP(app(filter, f''), app(app(cons, x), app(app(cons, x''), xs''))) -> APP(app(app(if, app(f'', x)), app(app(cons, x), app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs'')))), app(app(filter, f''), app(app(cons, x''), xs'')))
APP(app(filter, f''), app(app(cons, x), nil)) -> APP(app(app(if, app(f'', x)), app(app(cons, x), app(app(filter, f''), nil))), nil)
APP(app(filter, f''), app(app(cons, x), app(app(cons, x''), xs''))) -> APP(app(app(if, app(f'', x)), app(app(cons, x), app(app(filter, f''), app(app(cons, x''), xs'')))), app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs'')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Rewriting Transformation


Dependency Pairs:

APP(app(filter, f''), app(app(cons, x), app(app(cons, x''), xs''))) -> APP(app(app(if, app(f'', x)), app(app(cons, x), app(app(filter, f''), app(app(cons, x''), xs'')))), app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs'')))
APP(app(filter, f''), app(app(cons, x), app(app(cons, x''), xs''))) -> APP(app(app(if, app(f'', x)), app(app(cons, x), app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs'')))), app(app(filter, f''), app(app(cons, x''), xs'')))
APP(app(filter, f''), app(app(cons, x), nil)) -> APP(app(app(if, app(f'', x)), app(app(cons, x), nil)), app(app(filter, f''), nil))
APP(app(filter, app(filter, f'')), app(app(cons, app(app(cons, x''), xs'')), xs)) -> APP(app(app(if, app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs''))), app(app(cons, app(app(cons, x''), xs'')), app(app(filter, app(filter, f'')), xs))), app(app(filter, app(filter, f'')), xs))
APP(app(filter, app(filter, f'')), app(app(cons, nil), xs)) -> APP(app(app(if, nil), app(app(cons, nil), app(app(filter, app(filter, f'')), xs))), app(app(filter, app(filter, f'')), xs))
APP(app(filter, app(app(if, false), x'')), app(app(cons, x0), xs)) -> APP(app(app(if, x0), app(app(cons, x0), app(app(filter, app(app(if, false), x'')), xs))), app(app(filter, app(app(if, false), x'')), xs))
APP(app(filter, app(app(if, true), x'')), app(app(cons, x0), xs)) -> APP(app(app(if, x''), app(app(cons, x0), app(app(filter, app(app(if, true), x'')), xs))), app(app(filter, app(app(if, true), x'')), xs))
APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs)))
APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(filter, f), xs)


Rules:


app(app(app(if, true), x), y) -> x
app(app(app(if, false), x), y) -> y
app(app(filter, f), nil) -> nil
app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs))), app(app(filter, f), xs))


Strategy:

innermost




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

APP(app(filter, f''), app(app(cons, x), nil)) -> APP(app(app(if, app(f'', x)), app(app(cons, x), nil)), app(app(filter, f''), nil))
one new Dependency Pair is created:

APP(app(filter, f''), app(app(cons, x), nil)) -> APP(app(app(if, app(f'', x)), app(app(cons, x), nil)), nil)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Rw
             ...
               →DP Problem 3
Rewriting Transformation


Dependency Pairs:

APP(app(filter, f''), app(app(cons, x), app(app(cons, x''), xs''))) -> APP(app(app(if, app(f'', x)), app(app(cons, x), app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs'')))), app(app(filter, f''), app(app(cons, x''), xs'')))
APP(app(filter, app(filter, f'')), app(app(cons, app(app(cons, x''), xs'')), xs)) -> APP(app(app(if, app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs''))), app(app(cons, app(app(cons, x''), xs'')), app(app(filter, app(filter, f'')), xs))), app(app(filter, app(filter, f'')), xs))
APP(app(filter, app(filter, f'')), app(app(cons, nil), xs)) -> APP(app(app(if, nil), app(app(cons, nil), app(app(filter, app(filter, f'')), xs))), app(app(filter, app(filter, f'')), xs))
APP(app(filter, app(app(if, false), x'')), app(app(cons, x0), xs)) -> APP(app(app(if, x0), app(app(cons, x0), app(app(filter, app(app(if, false), x'')), xs))), app(app(filter, app(app(if, false), x'')), xs))
APP(app(filter, app(app(if, true), x'')), app(app(cons, x0), xs)) -> APP(app(app(if, x''), app(app(cons, x0), app(app(filter, app(app(if, true), x'')), xs))), app(app(filter, app(app(if, true), x'')), xs))
APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs)))
APP(app(filter, f''), app(app(cons, x), app(app(cons, x''), xs''))) -> APP(app(app(if, app(f'', x)), app(app(cons, x), app(app(filter, f''), app(app(cons, x''), xs'')))), app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs'')))


Rules:


app(app(app(if, true), x), y) -> x
app(app(app(if, false), x), y) -> y
app(app(filter, f), nil) -> nil
app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs))), app(app(filter, f), xs))


Strategy:

innermost




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

APP(app(filter, f''), app(app(cons, x), app(app(cons, x''), xs''))) -> APP(app(app(if, app(f'', x)), app(app(cons, x), app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs'')))), app(app(filter, f''), app(app(cons, x''), xs'')))
one new Dependency Pair is created:

APP(app(filter, f''), app(app(cons, x), app(app(cons, x''), xs''))) -> APP(app(app(if, app(f'', x)), app(app(cons, x), app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs'')))), app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs'')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Rw
             ...
               →DP Problem 4
Rewriting Transformation


Dependency Pairs:

APP(app(filter, f''), app(app(cons, x), app(app(cons, x''), xs''))) -> APP(app(app(if, app(f'', x)), app(app(cons, x), app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs'')))), app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs'')))
APP(app(filter, f''), app(app(cons, x), app(app(cons, x''), xs''))) -> APP(app(app(if, app(f'', x)), app(app(cons, x), app(app(filter, f''), app(app(cons, x''), xs'')))), app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs'')))
APP(app(filter, app(filter, f'')), app(app(cons, nil), xs)) -> APP(app(app(if, nil), app(app(cons, nil), app(app(filter, app(filter, f'')), xs))), app(app(filter, app(filter, f'')), xs))
APP(app(filter, app(app(if, false), x'')), app(app(cons, x0), xs)) -> APP(app(app(if, x0), app(app(cons, x0), app(app(filter, app(app(if, false), x'')), xs))), app(app(filter, app(app(if, false), x'')), xs))
APP(app(filter, app(app(if, true), x'')), app(app(cons, x0), xs)) -> APP(app(app(if, x''), app(app(cons, x0), app(app(filter, app(app(if, true), x'')), xs))), app(app(filter, app(app(if, true), x'')), xs))
APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs)))
APP(app(filter, app(filter, f'')), app(app(cons, app(app(cons, x''), xs'')), xs)) -> APP(app(app(if, app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs''))), app(app(cons, app(app(cons, x''), xs'')), app(app(filter, app(filter, f'')), xs))), app(app(filter, app(filter, f'')), xs))


Rules:


app(app(app(if, true), x), y) -> x
app(app(app(if, false), x), y) -> y
app(app(filter, f), nil) -> nil
app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs))), app(app(filter, f), xs))


Strategy:

innermost




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

APP(app(filter, f''), app(app(cons, x), app(app(cons, x''), xs''))) -> APP(app(app(if, app(f'', x)), app(app(cons, x), app(app(filter, f''), app(app(cons, x''), xs'')))), app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs'')))
one new Dependency Pair is created:

APP(app(filter, f''), app(app(cons, x), app(app(cons, x''), xs''))) -> APP(app(app(if, app(f'', x)), app(app(cons, x), app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs'')))), app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs'')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Rw
             ...
               →DP Problem 5
Narrowing Transformation


Dependency Pairs:

APP(app(filter, f''), app(app(cons, x), app(app(cons, x''), xs''))) -> APP(app(app(if, app(f'', x)), app(app(cons, x), app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs'')))), app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs'')))
APP(app(filter, app(filter, f'')), app(app(cons, app(app(cons, x''), xs'')), xs)) -> APP(app(app(if, app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs''))), app(app(cons, app(app(cons, x''), xs'')), app(app(filter, app(filter, f'')), xs))), app(app(filter, app(filter, f'')), xs))
APP(app(filter, app(filter, f'')), app(app(cons, nil), xs)) -> APP(app(app(if, nil), app(app(cons, nil), app(app(filter, app(filter, f'')), xs))), app(app(filter, app(filter, f'')), xs))
APP(app(filter, app(app(if, false), x'')), app(app(cons, x0), xs)) -> APP(app(app(if, x0), app(app(cons, x0), app(app(filter, app(app(if, false), x'')), xs))), app(app(filter, app(app(if, false), x'')), xs))
APP(app(filter, app(app(if, true), x'')), app(app(cons, x0), xs)) -> APP(app(app(if, x''), app(app(cons, x0), app(app(filter, app(app(if, true), x'')), xs))), app(app(filter, app(app(if, true), x'')), xs))
APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs)))
APP(app(filter, f''), app(app(cons, x), app(app(cons, x''), xs''))) -> APP(app(app(if, app(f'', x)), app(app(cons, x), app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs'')))), app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs'')))


Rules:


app(app(app(if, true), x), y) -> x
app(app(app(if, false), x), y) -> y
app(app(filter, f), nil) -> nil
app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs))), app(app(filter, f), xs))


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, x), xs)) -> APP(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs)))
six new Dependency Pairs are created:

APP(app(filter, app(app(if, true), x'')), app(app(cons, x0), xs)) -> APP(app(if, x''), app(app(cons, x0), app(app(filter, app(app(if, true), x'')), xs)))
APP(app(filter, app(app(if, false), x'')), app(app(cons, x0), xs)) -> APP(app(if, x0), app(app(cons, x0), app(app(filter, app(app(if, false), x'')), xs)))
APP(app(filter, app(filter, f'')), app(app(cons, nil), xs)) -> APP(app(if, nil), app(app(cons, nil), app(app(filter, app(filter, f'')), xs)))
APP(app(filter, app(filter, f'')), app(app(cons, app(app(cons, x''), xs'')), xs)) -> APP(app(if, app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs''))), app(app(cons, app(app(cons, x''), xs'')), app(app(filter, app(filter, f'')), xs)))
APP(app(filter, f''), app(app(cons, x), nil)) -> APP(app(if, app(f'', x)), app(app(cons, x), nil))
APP(app(filter, f''), app(app(cons, x), app(app(cons, x''), xs''))) -> APP(app(if, app(f'', x)), app(app(cons, x), app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs''))))

The transformation is resulting in one new DP problem:



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




The following remains to be proven:
Dependency Pairs:

APP(app(filter, f''), app(app(cons, x), app(app(cons, x''), xs''))) -> APP(app(if, app(f'', x)), app(app(cons, x), app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs''))))
APP(app(filter, f''), app(app(cons, x), nil)) -> APP(app(if, app(f'', x)), app(app(cons, x), nil))
APP(app(filter, app(filter, f'')), app(app(cons, app(app(cons, x''), xs'')), xs)) -> APP(app(if, app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs''))), app(app(cons, app(app(cons, x''), xs'')), app(app(filter, app(filter, f'')), xs)))
APP(app(filter, app(filter, f'')), app(app(cons, nil), xs)) -> APP(app(if, nil), app(app(cons, nil), app(app(filter, app(filter, f'')), xs)))
APP(app(filter, app(app(if, false), x'')), app(app(cons, x0), xs)) -> APP(app(if, x0), app(app(cons, x0), app(app(filter, app(app(if, false), x'')), xs)))
APP(app(filter, app(app(if, true), x'')), app(app(cons, x0), xs)) -> APP(app(if, x''), app(app(cons, x0), app(app(filter, app(app(if, true), x'')), xs)))
APP(app(filter, f''), app(app(cons, x), app(app(cons, x''), xs''))) -> APP(app(app(if, app(f'', x)), app(app(cons, x), app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs'')))), app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs'')))
APP(app(filter, app(filter, f'')), app(app(cons, app(app(cons, x''), xs'')), xs)) -> APP(app(app(if, app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs''))), app(app(cons, app(app(cons, x''), xs'')), app(app(filter, app(filter, f'')), xs))), app(app(filter, app(filter, f'')), xs))
APP(app(filter, app(filter, f'')), app(app(cons, nil), xs)) -> APP(app(app(if, nil), app(app(cons, nil), app(app(filter, app(filter, f'')), xs))), app(app(filter, app(filter, f'')), xs))
APP(app(filter, app(app(if, false), x'')), app(app(cons, x0), xs)) -> APP(app(app(if, x0), app(app(cons, x0), app(app(filter, app(app(if, false), x'')), xs))), app(app(filter, app(app(if, false), x'')), xs))
APP(app(filter, app(app(if, true), x'')), app(app(cons, x0), xs)) -> APP(app(app(if, x''), app(app(cons, x0), app(app(filter, app(app(if, true), x'')), xs))), app(app(filter, app(app(if, true), x'')), xs))
APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x)
APP(app(filter, f''), app(app(cons, x), app(app(cons, x''), xs''))) -> APP(app(app(if, app(f'', x)), app(app(cons, x), app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs'')))), app(app(app(if, app(f'', x'')), app(app(cons, x''), app(app(filter, f''), xs''))), app(app(filter, f''), xs'')))


Rules:


app(app(app(if, true), x), y) -> x
app(app(app(if, false), x), y) -> y
app(app(filter, f), nil) -> nil
app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs))), app(app(filter, f), xs))


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:28 minutes