Term Rewriting System R:
[l, h, t, f, l1, l2, l3]
app(app(append, nil), l) -> l
app(app(append, app(app(cons, h), t)), l) -> app(app(cons, h), app(app(append, t), l))
app(app(map, f), nil) -> nil
app(app(map, f), app(app(cons, h), t)) -> app(app(cons, app(f, h)), app(app(map, f), t))
app(app(append, app(app(append, l1), l2)), l3) -> app(app(append, l1), app(app(append, l2), l3))
app(app(map, f), app(app(append, l1), l2)) -> app(app(append, app(app(map, f), l1)), app(app(map, f), l2))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

APP(app(append, app(app(cons, h), t)), l) -> APP(app(cons, h), app(app(append, t), l))
APP(app(append, app(app(cons, h), t)), l) -> APP(app(append, t), l)
APP(app(append, app(app(cons, h), t)), l) -> APP(append, t)
APP(app(map, f), app(app(cons, h), t)) -> APP(app(cons, app(f, h)), app(app(map, f), t))
APP(app(map, f), app(app(cons, h), t)) -> APP(cons, app(f, h))
APP(app(map, f), app(app(cons, h), t)) -> APP(f, h)
APP(app(map, f), app(app(cons, h), t)) -> APP(app(map, f), t)
APP(app(append, app(app(append, l1), l2)), l3) -> APP(app(append, l1), app(app(append, l2), l3))
APP(app(append, app(app(append, l1), l2)), l3) -> APP(app(append, l2), l3)
APP(app(append, app(app(append, l1), l2)), l3) -> APP(append, l2)
APP(app(map, f), app(app(append, l1), l2)) -> APP(app(append, app(app(map, f), l1)), app(app(map, f), l2))
APP(app(map, f), app(app(append, l1), l2)) -> APP(append, app(app(map, f), l1))
APP(app(map, f), app(app(append, l1), l2)) -> APP(app(map, f), l1)
APP(app(map, f), app(app(append, l1), l2)) -> APP(app(map, f), l2)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation


Dependency Pairs:

APP(app(map, f), app(app(append, l1), l2)) -> APP(app(map, f), l2)
APP(app(map, f), app(app(append, l1), l2)) -> APP(app(map, f), l1)
APP(app(map, f), app(app(append, l1), l2)) -> APP(app(append, app(app(map, f), l1)), app(app(map, f), l2))
APP(app(append, app(app(append, l1), l2)), l3) -> APP(app(append, l2), l3)
APP(app(append, app(app(append, l1), l2)), l3) -> APP(app(append, l1), app(app(append, l2), l3))
APP(app(map, f), app(app(cons, h), t)) -> APP(app(map, f), t)
APP(app(map, f), app(app(cons, h), t)) -> APP(f, h)
APP(app(map, f), app(app(cons, h), t)) -> APP(app(cons, app(f, h)), app(app(map, f), t))
APP(app(append, app(app(cons, h), t)), l) -> APP(app(append, t), l)
APP(app(append, app(app(cons, h), t)), l) -> APP(app(cons, h), app(app(append, t), l))


Rules:


app(app(append, nil), l) -> l
app(app(append, app(app(cons, h), t)), l) -> app(app(cons, h), app(app(append, t), l))
app(app(map, f), nil) -> nil
app(app(map, f), app(app(cons, h), t)) -> app(app(cons, app(f, h)), app(app(map, f), t))
app(app(append, app(app(append, l1), l2)), l3) -> app(app(append, l1), app(app(append, l2), l3))
app(app(map, f), app(app(append, l1), l2)) -> app(app(append, app(app(map, f), l1)), app(app(map, f), l2))





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

APP(app(append, app(app(cons, h), t)), l) -> APP(app(cons, h), app(app(append, t), l))
three new Dependency Pairs are created:

APP(app(append, app(app(cons, h), nil)), l'') -> APP(app(cons, h), l'')
APP(app(append, app(app(cons, h), app(app(cons, h''), t''))), l'') -> APP(app(cons, h), app(app(cons, h''), app(app(append, t''), l'')))
APP(app(append, app(app(cons, h), app(app(append, l1'), l2'))), l') -> APP(app(cons, h), app(app(append, l1'), app(app(append, l2'), l')))

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

APP(app(append, app(app(cons, h), app(app(append, l1'), l2'))), l') -> APP(app(cons, h), app(app(append, l1'), app(app(append, l2'), l')))
APP(app(append, app(app(cons, h), app(app(cons, h''), t''))), l'') -> APP(app(cons, h), app(app(cons, h''), app(app(append, t''), l'')))
APP(app(append, app(app(cons, h), nil)), l'') -> APP(app(cons, h), l'')
APP(app(map, f), app(app(append, l1), l2)) -> APP(app(map, f), l1)
APP(app(map, f), app(app(append, l1), l2)) -> APP(app(append, app(app(map, f), l1)), app(app(map, f), l2))
APP(app(append, app(app(append, l1), l2)), l3) -> APP(app(append, l2), l3)
APP(app(append, app(app(append, l1), l2)), l3) -> APP(app(append, l1), app(app(append, l2), l3))
APP(app(map, f), app(app(cons, h), t)) -> APP(app(map, f), t)
APP(app(map, f), app(app(cons, h), t)) -> APP(f, h)
APP(app(map, f), app(app(cons, h), t)) -> APP(app(cons, app(f, h)), app(app(map, f), t))
APP(app(append, app(app(cons, h), t)), l) -> APP(app(append, t), l)
APP(app(map, f), app(app(append, l1), l2)) -> APP(app(map, f), l2)


Rules:


app(app(append, nil), l) -> l
app(app(append, app(app(cons, h), t)), l) -> app(app(cons, h), app(app(append, t), l))
app(app(map, f), nil) -> nil
app(app(map, f), app(app(cons, h), t)) -> app(app(cons, app(f, h)), app(app(map, f), t))
app(app(append, app(app(append, l1), l2)), l3) -> app(app(append, l1), app(app(append, l2), l3))
app(app(map, f), app(app(append, l1), l2)) -> app(app(append, app(app(map, f), l1)), app(app(map, f), l2))





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

APP(app(append, app(app(cons, h), nil)), l'') -> APP(app(cons, h), l'')
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 3
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

APP(app(append, app(app(cons, h), app(app(cons, h''), t''))), l'') -> APP(app(cons, h), app(app(cons, h''), app(app(append, t''), l'')))
APP(app(map, f), app(app(append, l1), l2)) -> APP(app(map, f), l2)
APP(app(map, f), app(app(append, l1), l2)) -> APP(app(map, f), l1)
APP(app(map, f), app(app(append, l1), l2)) -> APP(app(append, app(app(map, f), l1)), app(app(map, f), l2))
APP(app(append, app(app(append, l1), l2)), l3) -> APP(app(append, l2), l3)
APP(app(append, app(app(append, l1), l2)), l3) -> APP(app(append, l1), app(app(append, l2), l3))
APP(app(map, f), app(app(cons, h), t)) -> APP(app(map, f), t)
APP(app(map, f), app(app(cons, h), t)) -> APP(f, h)
APP(app(map, f), app(app(cons, h), t)) -> APP(app(cons, app(f, h)), app(app(map, f), t))
APP(app(append, app(app(cons, h), t)), l) -> APP(app(append, t), l)
APP(app(append, app(app(cons, h), app(app(append, l1'), l2'))), l') -> APP(app(cons, h), app(app(append, l1'), app(app(append, l2'), l')))


Rules:


app(app(append, nil), l) -> l
app(app(append, app(app(cons, h), t)), l) -> app(app(cons, h), app(app(append, t), l))
app(app(map, f), nil) -> nil
app(app(map, f), app(app(cons, h), t)) -> app(app(cons, app(f, h)), app(app(map, f), t))
app(app(append, app(app(append, l1), l2)), l3) -> app(app(append, l1), app(app(append, l2), l3))
app(app(map, f), app(app(append, l1), l2)) -> app(app(append, app(app(map, f), l1)), app(app(map, f), l2))




Termination of R could not be shown.
Duration:
0:01 minutes