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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

APP(app(minus, x), app(s, y)) -> APP(pred, app(app(minus, x), y))
APP(app(minus, x), app(s, y)) -> APP(app(minus, x), y)
APP(app(quot, app(s, x)), app(s, y)) -> APP(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
APP(app(quot, app(s, x)), app(s, y)) -> APP(app(quot, app(app(minus, x), y)), app(s, y))
APP(app(quot, app(s, x)), app(s, y)) -> APP(quot, app(app(minus, x), y))
APP(app(quot, app(s, x)), app(s, y)) -> APP(app(minus, x), y)
APP(app(quot, app(s, x)), app(s, y)) -> APP(minus, x)

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Nar


Dependency Pair:

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


Rules:


app(pred, app(s, x)) -> x
app(app(minus, x), 0) -> x
app(app(minus, x), app(s, y)) -> app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) -> 0
app(app(quot, app(s, x)), app(s, y)) -> app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(pred)=  0  
  POL(minus)=  0  
  POL(quot)=  0  
  POL(s)=  1  
  POL(app(x1, x2))=  x1 + x2  
  POL(APP(x1, x2))=  x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 3
             ↳Dependency Graph
       →DP Problem 2
         ↳Nar


Dependency Pair:


Rules:


app(pred, app(s, x)) -> x
app(app(minus, x), 0) -> x
app(app(minus, x), app(s, y)) -> app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) -> 0
app(app(quot, app(s, x)), app(s, y)) -> app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pairs:

APP(app(quot, app(s, x)), app(s, y)) -> APP(app(minus, x), y)
APP(app(quot, app(s, x)), app(s, y)) -> APP(app(quot, app(app(minus, x), y)), app(s, y))


Rules:


app(pred, app(s, x)) -> x
app(app(minus, x), 0) -> x
app(app(minus, x), app(s, y)) -> app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) -> 0
app(app(quot, app(s, x)), app(s, y)) -> app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))


Strategy:

innermost




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

APP(app(quot, app(s, x)), app(s, y)) -> APP(app(minus, x), y)
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pair:

APP(app(quot, app(s, x)), app(s, y)) -> APP(app(quot, app(app(minus, x), y)), app(s, y))


Rules:


app(pred, app(s, x)) -> x
app(app(minus, x), 0) -> x
app(app(minus, x), app(s, y)) -> app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) -> 0
app(app(quot, app(s, x)), app(s, y)) -> app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))


Strategy:

innermost



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