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))

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(app(cons, x), 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))





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(cons, x), app(app(filter, f), xs))
two new Dependency Pairs are created:

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




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