Term Rewriting System R:
[y, z, x]
app(f, app(app(cons, nil), y)) -> y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) -> app(f, z)
app(app(app(copy, app(s, x)), y), z) -> app(app(app(copy, x), y), app(app(cons, app(f, y)), z))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> APP(app(app(copy, n), y), z)
APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> APP(app(copy, n), y)
APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> APP(copy, n)
APP(app(app(copy, 0), y), z) -> APP(f, z)
APP(app(app(copy, app(s, x)), y), z) -> APP(app(app(copy, x), y), app(app(cons, app(f, y)), z))
APP(app(app(copy, app(s, x)), y), z) -> APP(app(copy, x), y)
APP(app(app(copy, app(s, x)), y), z) -> APP(copy, x)
APP(app(app(copy, app(s, x)), y), z) -> APP(app(cons, app(f, y)), z)
APP(app(app(copy, app(s, x)), y), z) -> APP(cons, app(f, y))
APP(app(app(copy, app(s, x)), y), z) -> APP(f, y)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation


Dependency Pairs:

APP(app(app(copy, app(s, x)), y), z) -> APP(f, y)
APP(app(app(copy, app(s, x)), y), z) -> APP(app(cons, app(f, y)), z)
APP(app(app(copy, app(s, x)), y), z) -> APP(app(copy, x), y)
APP(app(app(copy, app(s, x)), y), z) -> APP(app(app(copy, x), y), app(app(cons, app(f, y)), z))
APP(app(app(copy, 0), y), z) -> APP(f, z)
APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> APP(app(copy, n), y)
APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> APP(app(app(copy, n), y), z)


Rules:


app(f, app(app(cons, nil), y)) -> y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) -> app(f, z)
app(app(app(copy, app(s, x)), y), z) -> app(app(app(copy, x), y), app(app(cons, app(f, y)), z))





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

APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> APP(app(app(copy, n), y), z)
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

APP(app(app(copy, app(s, x)), y), z) -> APP(app(cons, app(f, y)), z)
APP(app(app(copy, app(s, x)), y), z) -> APP(app(copy, x), y)
APP(app(app(copy, app(s, x)), y), z) -> APP(app(app(copy, x), y), app(app(cons, app(f, y)), z))
APP(app(app(copy, 0), y), z) -> APP(f, z)
APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> APP(app(copy, n), y)
APP(app(app(copy, app(s, x)), y), z) -> APP(f, y)


Rules:


app(f, app(app(cons, nil), y)) -> y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) -> app(f, z)
app(app(app(copy, app(s, x)), y), z) -> app(app(app(copy, x), y), app(app(cons, app(f, y)), z))





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

APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> APP(app(copy, n), 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 3
                 ↳Narrowing Transformation


Dependency Pairs:

APP(app(app(copy, app(s, x)), y), z) -> APP(app(copy, x), y)
APP(app(app(copy, app(s, x)), y), z) -> APP(app(app(copy, x), y), app(app(cons, app(f, y)), z))
APP(app(app(copy, app(s, x)), y), z) -> APP(app(cons, app(f, y)), z)


Rules:


app(f, app(app(cons, nil), y)) -> y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) -> app(f, z)
app(app(app(copy, app(s, x)), y), z) -> app(app(app(copy, x), y), app(app(cons, app(f, y)), z))





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

APP(app(app(copy, app(s, x)), y), z) -> APP(app(copy, x), 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(app(copy, app(s, x)), y), z) -> APP(app(cons, app(f, y)), z)
APP(app(app(copy, app(s, x)), y), z) -> APP(app(app(copy, x), y), app(app(cons, app(f, y)), z))


Rules:


app(f, app(app(cons, nil), y)) -> y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) -> app(f, z)
app(app(app(copy, app(s, x)), y), z) -> app(app(app(copy, x), y), app(app(cons, app(f, y)), z))




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