Term Rewriting System R:
[x, y, z]
app(app(., 1), x) -> x
app(app(., x), 1) -> x
app(app(., app(i, x)), x) -> 1
app(app(., x), app(i, x)) -> 1
app(app(., app(i, y)), app(app(., y), z)) -> z
app(app(., y), app(app(., app(i, y)), z)) -> z
app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))
app(i, 1) -> 1
app(i, app(i, x)) -> x
app(i, app(app(., x), y)) -> app(app(., app(i, y)), app(i, x))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

APP(app(., app(app(., x), y)), z) -> APP(app(., x), app(app(., y), z))
APP(app(., app(app(., x), y)), z) -> APP(app(., y), z)
APP(app(., app(app(., x), y)), z) -> APP(., y)
APP(i, app(app(., x), y)) -> APP(app(., app(i, y)), app(i, x))
APP(i, app(app(., x), y)) -> APP(., app(i, y))
APP(i, app(app(., x), y)) -> APP(i, y)
APP(i, app(app(., x), y)) -> APP(i, x)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering


Dependency Pairs:

APP(i, app(app(., x), y)) -> APP(i, x)
APP(i, app(app(., x), y)) -> APP(i, y)
APP(i, app(app(., x), y)) -> APP(app(., app(i, y)), app(i, x))
APP(app(., app(app(., x), y)), z) -> APP(app(., y), z)
APP(app(., app(app(., x), y)), z) -> APP(app(., x), app(app(., y), z))


Rules:


app(app(., 1), x) -> x
app(app(., x), 1) -> x
app(app(., app(i, x)), x) -> 1
app(app(., x), app(i, x)) -> 1
app(app(., app(i, y)), app(app(., y), z)) -> z
app(app(., y), app(app(., app(i, y)), z)) -> z
app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))
app(i, 1) -> 1
app(i, app(i, x)) -> x
app(i, app(app(., x), y)) -> app(app(., app(i, y)), app(i, x))





The following dependency pairs can be strictly oriented:

APP(i, app(app(., x), y)) -> APP(i, x)
APP(i, app(app(., x), y)) -> APP(i, y)
APP(app(., app(app(., x), y)), z) -> APP(app(., y), z)


Additionally, the following rules can be oriented:

app(app(., 1), x) -> x
app(app(., x), 1) -> x
app(app(., app(i, x)), x) -> 1
app(app(., x), app(i, x)) -> 1
app(app(., app(i, y)), app(app(., y), z)) -> z
app(app(., y), app(app(., app(i, y)), z)) -> z
app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))
app(i, 1) -> 1
app(i, app(i, x)) -> x
app(i, app(app(., x), y)) -> app(app(., app(i, y)), app(i, x))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(i)=  0  
  POL(1)=  0  
  POL(.)=  1  
  POL(app(x1, x2))=  x1 + x2  
  POL(APP(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

APP(i, app(app(., x), y)) -> APP(app(., app(i, y)), app(i, x))
APP(app(., app(app(., x), y)), z) -> APP(app(., x), app(app(., y), z))


Rules:


app(app(., 1), x) -> x
app(app(., x), 1) -> x
app(app(., app(i, x)), x) -> 1
app(app(., x), app(i, x)) -> 1
app(app(., app(i, y)), app(app(., y), z)) -> z
app(app(., y), app(app(., app(i, y)), z)) -> z
app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))
app(i, 1) -> 1
app(i, app(i, x)) -> x
app(i, app(app(., x), y)) -> app(app(., app(i, y)), app(i, x))




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