Term Rewriting System R:
[y, x, g, f, xs]
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(times, 0), y) -> 0
app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
app(app(app(curry, g), x), y) -> app(app(g, x), y)
app(app(map, f), nil) -> nil
app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs))
inc -> app(map, app(app(curry, plus), app(s, 0)))
double -> app(map, app(app(curry, times), app(s, app(s, 0))))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

APP(app(plus, app(s, x)), y) -> APP(s, app(app(plus, x), y))
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(plus, app(s, x)), y) -> APP(plus, x)
APP(app(times, app(s, x)), y) -> APP(app(plus, app(app(times, x), y)), y)
APP(app(times, app(s, x)), y) -> APP(plus, app(app(times, x), y))
APP(app(times, app(s, x)), y) -> APP(app(times, x), y)
APP(app(times, app(s, x)), y) -> APP(times, x)
APP(app(app(curry, g), x), y) -> APP(app(g, x), y)
APP(app(app(curry, g), x), y) -> APP(g, x)
APP(app(map, f), app(app(cons, x), xs)) -> APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(map, f), app(app(cons, x), xs)) -> APP(cons, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)
INC -> APP(map, app(app(curry, plus), app(s, 0)))
INC -> APP(app(curry, plus), app(s, 0))
INC -> APP(curry, plus)
INC -> APP(s, 0)
DOUBLE -> APP(map, app(app(curry, times), app(s, app(s, 0))))
DOUBLE -> APP(app(curry, times), app(s, app(s, 0)))
DOUBLE -> APP(curry, times)
DOUBLE -> APP(s, app(s, 0))
DOUBLE -> APP(s, 0)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation


Dependency Pairs:

APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)
APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) -> APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(app(curry, g), x), y) -> APP(g, x)
APP(app(app(curry, g), x), y) -> APP(app(g, x), y)
APP(app(times, app(s, x)), y) -> APP(app(times, x), y)
APP(app(times, app(s, x)), y) -> APP(app(plus, app(app(times, x), y)), y)
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)


Rules:


app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(times, 0), y) -> 0
app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
app(app(app(curry, g), x), y) -> app(app(g, x), y)
app(app(map, f), nil) -> nil
app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs))
inc -> app(map, app(app(curry, plus), app(s, 0)))
double -> app(map, app(app(curry, times), app(s, app(s, 0))))


Strategy:

innermost




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

APP(app(times, app(s, x)), y) -> APP(app(plus, app(app(times, x), y)), y)
two new Dependency Pairs are created:

APP(app(times, app(s, 0)), y'') -> APP(app(plus, 0), y'')
APP(app(times, app(s, app(s, x''))), y'') -> APP(app(plus, app(app(plus, app(app(times, x''), y'')), y'')), y'')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Narrowing Transformation


Dependency Pairs:

APP(app(times, app(s, app(s, x''))), y'') -> APP(app(plus, app(app(plus, app(app(times, x''), y'')), y'')), y'')
APP(app(times, app(s, 0)), y'') -> APP(app(plus, 0), y'')
APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)
APP(app(app(curry, g), x), y) -> APP(g, x)
APP(app(app(curry, g), x), y) -> APP(app(g, x), y)
APP(app(times, app(s, x)), y) -> APP(app(times, x), y)
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(map, f), app(app(cons, x), xs)) -> APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)


Rules:


app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(times, 0), y) -> 0
app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
app(app(app(curry, g), x), y) -> app(app(g, x), y)
app(app(map, f), nil) -> nil
app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs))
inc -> app(map, app(app(curry, plus), app(s, 0)))
double -> app(map, app(app(curry, times), app(s, app(s, 0))))


Strategy:

innermost




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

APP(app(map, f), app(app(cons, x), xs)) -> APP(app(cons, app(f, x)), app(app(map, f), xs))
nine new Dependency Pairs are created:

APP(app(map, app(plus, 0)), app(app(cons, x'), xs)) -> APP(app(cons, x'), app(app(map, app(plus, 0)), xs))
APP(app(map, app(plus, app(s, x''))), app(app(cons, x0), xs)) -> APP(app(cons, app(s, app(app(plus, x''), x0))), app(app(map, app(plus, app(s, x''))), xs))
APP(app(map, app(times, 0)), app(app(cons, x'), xs)) -> APP(app(cons, 0), app(app(map, app(times, 0)), xs))
APP(app(map, app(times, app(s, x''))), app(app(cons, x0), xs)) -> APP(app(cons, app(app(plus, app(app(times, x''), x0)), x0)), app(app(map, app(times, app(s, x''))), xs))
APP(app(map, app(app(curry, g'), x'')), app(app(cons, x0), xs)) -> APP(app(cons, app(app(g', x''), x0)), app(app(map, app(app(curry, g'), x'')), xs))
APP(app(map, app(map, f'')), app(app(cons, nil), xs)) -> APP(app(cons, nil), app(app(map, app(map, f'')), xs))
APP(app(map, app(map, f'')), app(app(cons, app(app(cons, x''), xs'')), xs)) -> APP(app(cons, app(app(cons, app(f'', x'')), app(app(map, f''), xs''))), app(app(map, app(map, f'')), xs))
APP(app(map, f''), app(app(cons, x), nil)) -> APP(app(cons, app(f'', x)), nil)
APP(app(map, f''), app(app(cons, x), app(app(cons, x''), xs''))) -> APP(app(cons, app(f'', x)), app(app(cons, app(f'', x'')), app(app(map, f''), xs'')))

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(map, f''), app(app(cons, x), app(app(cons, x''), xs''))) -> APP(app(cons, app(f'', x)), app(app(cons, app(f'', x'')), app(app(map, f''), xs'')))
APP(app(map, f''), app(app(cons, x), nil)) -> APP(app(cons, app(f'', x)), nil)
APP(app(map, app(map, f'')), app(app(cons, app(app(cons, x''), xs'')), xs)) -> APP(app(cons, app(app(cons, app(f'', x'')), app(app(map, f''), xs''))), app(app(map, app(map, f'')), xs))
APP(app(map, app(app(curry, g'), x'')), app(app(cons, x0), xs)) -> APP(app(cons, app(app(g', x''), x0)), app(app(map, app(app(curry, g'), x'')), xs))
APP(app(map, app(times, app(s, x''))), app(app(cons, x0), xs)) -> APP(app(cons, app(app(plus, app(app(times, x''), x0)), x0)), app(app(map, app(times, app(s, x''))), xs))
APP(app(map, app(times, 0)), app(app(cons, x'), xs)) -> APP(app(cons, 0), app(app(map, app(times, 0)), xs))
APP(app(times, app(s, 0)), y'') -> APP(app(plus, 0), y'')
APP(app(map, app(plus, app(s, x''))), app(app(cons, x0), xs)) -> APP(app(cons, app(s, app(app(plus, x''), x0))), app(app(map, app(plus, app(s, x''))), xs))
APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)
APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)
APP(app(app(curry, g), x), y) -> APP(g, x)
APP(app(app(curry, g), x), y) -> APP(app(g, x), y)
APP(app(times, app(s, x)), y) -> APP(app(times, x), y)
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(times, app(s, app(s, x''))), y'') -> APP(app(plus, app(app(plus, app(app(times, x''), y'')), y'')), y'')


Rules:


app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(times, 0), y) -> 0
app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
app(app(app(curry, g), x), y) -> app(app(g, x), y)
app(app(map, f), nil) -> nil
app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs))
inc -> app(map, app(app(curry, plus), app(s, 0)))
double -> app(map, app(app(curry, times), app(s, app(s, 0))))


Strategy:

innermost




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

APP(app(times, app(s, 0)), y'') -> APP(app(plus, 0), y'')
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
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

APP(app(map, f''), app(app(cons, x), nil)) -> APP(app(cons, app(f'', x)), nil)
APP(app(map, app(map, f'')), app(app(cons, app(app(cons, x''), xs'')), xs)) -> APP(app(cons, app(app(cons, app(f'', x'')), app(app(map, f''), xs''))), app(app(map, app(map, f'')), xs))
APP(app(map, app(app(curry, g'), x'')), app(app(cons, x0), xs)) -> APP(app(cons, app(app(g', x''), x0)), app(app(map, app(app(curry, g'), x'')), xs))
APP(app(map, app(times, app(s, x''))), app(app(cons, x0), xs)) -> APP(app(cons, app(app(plus, app(app(times, x''), x0)), x0)), app(app(map, app(times, app(s, x''))), xs))
APP(app(map, app(times, 0)), app(app(cons, x'), xs)) -> APP(app(cons, 0), app(app(map, app(times, 0)), xs))
APP(app(times, app(s, app(s, x''))), y'') -> APP(app(plus, app(app(plus, app(app(times, x''), y'')), y'')), y'')
APP(app(map, app(plus, app(s, x''))), app(app(cons, x0), xs)) -> APP(app(cons, app(s, app(app(plus, x''), x0))), app(app(map, app(plus, app(s, x''))), xs))
APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)
APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)
APP(app(app(curry, g), x), y) -> APP(g, x)
APP(app(app(curry, g), x), y) -> APP(app(g, x), y)
APP(app(times, app(s, x)), y) -> APP(app(times, x), y)
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(map, f''), app(app(cons, x), app(app(cons, x''), xs''))) -> APP(app(cons, app(f'', x)), app(app(cons, app(f'', x'')), app(app(map, f''), xs'')))


Rules:


app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(times, 0), y) -> 0
app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
app(app(app(curry, g), x), y) -> app(app(g, x), y)
app(app(map, f), nil) -> nil
app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs))
inc -> app(map, app(app(curry, plus), app(s, 0)))
double -> app(map, app(app(curry, times), app(s, app(s, 0))))


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:50 minutes