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(double, 0) -> 0
app(double, app(s, x)) -> app(s, app(s, app(double, x)))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) -> app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, app(app(minus, x), y)), app(double, y)))

Innermost 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(double, app(s, x)) -> APP(s, app(s, app(double, x)))
APP(double, app(s, x)) -> APP(s, app(double, x))
APP(double, app(s, x)) -> APP(double, x)
APP(app(plus, app(s, x)), y) -> APP(s, app(app(plus, x), y))
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(plus, app(s, x)), y) -> APP(plus, x)
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), app(s, y))
APP(app(plus, app(s, x)), y) -> APP(s, y)
APP(app(plus, app(s, x)), y) -> APP(s, app(app(plus, app(app(minus, x), y)), app(double, y)))
APP(app(plus, app(s, x)), y) -> APP(app(plus, app(app(minus, x), y)), app(double, y))
APP(app(plus, app(s, x)), y) -> APP(plus, app(app(minus, x), y))
APP(app(plus, app(s, x)), y) -> APP(app(minus, x), y)
APP(app(plus, app(s, x)), y) -> APP(minus, x)
APP(app(plus, app(s, x)), y) -> APP(double, y)

Furthermore, R contains two SCCs. 


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


Dependency Pair:

APP(double, app(s, x)) -> APP(double, x)


Rules:


app(app(minus, x), 0) -> x
app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)
app(double, 0) -> 0
app(double, app(s, x)) -> app(s, app(s, app(double, x)))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) -> app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, app(app(minus, x), y)), app(double, y)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

APP(double, app(s, x)) -> APP(double, x)


There are no usable rules for innermost that need to be oriented.

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

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


app(app(minus, x), 0) -> x
app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)
app(double, 0) -> 0
app(double, app(s, x)) -> app(s, app(s, app(double, x)))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) -> app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, app(app(minus, x), y)), app(double, y)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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




The following remains to be proven:
Dependency Pairs:

APP(app(plus, app(s, x)), y) -> APP(app(minus, x), y)
APP(app(plus, app(s, x)), y) -> APP(app(plus, app(app(minus, x), y)), app(double, y))
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), app(s, y))
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
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(double, 0) -> 0
app(double, app(s, x)) -> app(s, app(s, app(double, x)))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) -> app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, app(app(minus, x), y)), app(double, y)))


Strategy:

innermost



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