Term Rewriting System R:
[x, y]
app(app(minus, x), 0) -> x
app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)
app(f, 0) -> app(s, 0)
app(f, app(s, x)) -> app(app(minus, app(s, x)), app(g, app(f, x)))
app(g, 0) -> 0
app(g, app(s, x)) -> app(app(minus, app(s, x)), app(f, app(g, x)))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

APP(app(minus, app(s, x)), app(s, y)) -> APP(app(minus, x), y)
APP(app(minus, app(s, x)), app(s, y)) -> APP(minus, x)
APP(f, 0) -> APP(s, 0)
APP(f, app(s, x)) -> APP(app(minus, app(s, x)), app(g, app(f, x)))
APP(f, app(s, x)) -> APP(minus, app(s, x))
APP(f, app(s, x)) -> APP(g, app(f, x))
APP(f, app(s, x)) -> APP(f, x)
APP(g, app(s, x)) -> APP(app(minus, app(s, x)), app(f, app(g, x)))
APP(g, app(s, x)) -> APP(minus, app(s, x))
APP(g, app(s, x)) -> APP(f, app(g, x))
APP(g, app(s, x)) -> APP(g, x)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation


Dependency Pairs:

APP(g, app(s, x)) -> APP(g, x)
APP(g, app(s, x)) -> APP(f, app(g, x))
APP(f, app(s, x)) -> APP(f, x)
APP(g, app(s, x)) -> APP(app(minus, app(s, x)), app(f, app(g, x)))
APP(f, app(s, x)) -> APP(g, app(f, x))
APP(f, app(s, x)) -> APP(app(minus, app(s, x)), app(g, app(f, x)))
APP(app(minus, app(s, x)), app(s, y)) -> APP(app(minus, x), y)


Rules:


app(app(minus, x), 0) -> x
app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)
app(f, 0) -> app(s, 0)
app(f, app(s, x)) -> app(app(minus, app(s, x)), app(g, app(f, x)))
app(g, 0) -> 0
app(g, app(s, x)) -> app(app(minus, app(s, x)), app(f, app(g, x)))





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

APP(f, app(s, x)) -> APP(g, app(f, x))
two new Dependency Pairs are created:

APP(f, app(s, 0)) -> APP(g, app(s, 0))
APP(f, app(s, app(s, x''))) -> APP(g, app(app(minus, app(s, x'')), app(g, app(f, x''))))

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

APP(f, app(s, app(s, x''))) -> APP(g, app(app(minus, app(s, x'')), app(g, app(f, x''))))
APP(g, app(s, x)) -> APP(f, app(g, x))
APP(f, app(s, 0)) -> APP(g, app(s, 0))
APP(f, app(s, x)) -> APP(f, x)
APP(f, app(s, x)) -> APP(app(minus, app(s, x)), app(g, app(f, x)))
APP(app(minus, app(s, x)), app(s, y)) -> APP(app(minus, x), y)
APP(g, app(s, x)) -> APP(app(minus, app(s, x)), app(f, app(g, x)))
APP(g, app(s, x)) -> APP(g, x)


Rules:


app(app(minus, x), 0) -> x
app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)
app(f, 0) -> app(s, 0)
app(f, app(s, x)) -> app(app(minus, app(s, x)), app(g, app(f, x)))
app(g, 0) -> 0
app(g, app(s, x)) -> app(app(minus, app(s, x)), app(f, app(g, x)))





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

APP(g, app(s, x)) -> APP(f, app(g, x))
two new Dependency Pairs are created:

APP(g, app(s, 0)) -> APP(f, 0)
APP(g, app(s, app(s, x''))) -> APP(f, app(app(minus, app(s, x'')), app(f, app(g, x''))))

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(g, app(s, app(s, x''))) -> APP(f, app(app(minus, app(s, x'')), app(f, app(g, x''))))
APP(g, app(s, x)) -> APP(g, x)
APP(f, app(s, 0)) -> APP(g, app(s, 0))
APP(f, app(s, x)) -> APP(f, x)
APP(f, app(s, x)) -> APP(app(minus, app(s, x)), app(g, app(f, x)))
APP(app(minus, app(s, x)), app(s, y)) -> APP(app(minus, x), y)
APP(g, app(s, x)) -> APP(app(minus, app(s, x)), app(f, app(g, x)))
APP(f, app(s, app(s, x''))) -> APP(g, app(app(minus, app(s, x'')), app(g, app(f, x''))))


Rules:


app(app(minus, x), 0) -> x
app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)
app(f, 0) -> app(s, 0)
app(f, app(s, x)) -> app(app(minus, app(s, x)), app(g, app(f, x)))
app(g, 0) -> 0
app(g, app(s, x)) -> app(app(minus, app(s, x)), app(f, app(g, x)))




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