Term Rewriting System R:
[x, y]
app(D, t) -> 1
app(D, constant) -> 0
app(D, app(app(+, x), y)) -> app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) -> app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) -> app(app(-, app(D, x)), app(D, y))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

APP(D, app(app(+, x), y)) -> APP(app(+, app(D, x)), app(D, y))
APP(D, app(app(+, x), y)) -> APP(+, app(D, x))
APP(D, app(app(+, x), y)) -> APP(D, x)
APP(D, app(app(+, x), y)) -> APP(D, y)
APP(D, app(app(*, x), y)) -> APP(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
APP(D, app(app(*, x), y)) -> APP(+, app(app(*, y), app(D, x)))
APP(D, app(app(*, x), y)) -> APP(app(*, y), app(D, x))
APP(D, app(app(*, x), y)) -> APP(*, y)
APP(D, app(app(*, x), y)) -> APP(D, x)
APP(D, app(app(*, x), y)) -> APP(app(*, x), app(D, y))
APP(D, app(app(*, x), y)) -> APP(D, y)
APP(D, app(app(-, x), y)) -> APP(app(-, app(D, x)), app(D, y))
APP(D, app(app(-, x), y)) -> APP(-, app(D, x))
APP(D, app(app(-, x), y)) -> APP(D, x)
APP(D, app(app(-, x), y)) -> APP(D, y)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering


Dependency Pairs:

APP(D, app(app(-, x), y)) -> APP(D, y)
APP(D, app(app(-, x), y)) -> APP(D, x)
APP(D, app(app(-, x), y)) -> APP(app(-, app(D, x)), app(D, y))
APP(D, app(app(*, x), y)) -> APP(D, y)
APP(D, app(app(*, x), y)) -> APP(app(*, x), app(D, y))
APP(D, app(app(*, x), y)) -> APP(D, x)
APP(D, app(app(*, x), y)) -> APP(app(*, y), app(D, x))
APP(D, app(app(*, x), y)) -> APP(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
APP(D, app(app(+, x), y)) -> APP(D, y)
APP(D, app(app(+, x), y)) -> APP(D, x)
APP(D, app(app(+, x), y)) -> APP(app(+, app(D, x)), app(D, y))


Rules:


app(D, t) -> 1
app(D, constant) -> 0
app(D, app(app(+, x), y)) -> app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) -> app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) -> app(app(-, app(D, x)), app(D, y))





The following dependency pairs can be strictly oriented:

APP(D, app(app(-, x), y)) -> APP(app(-, app(D, x)), app(D, y))
APP(D, app(app(*, x), y)) -> APP(app(*, x), app(D, y))
APP(D, app(app(*, x), y)) -> APP(app(*, y), app(D, x))
APP(D, app(app(*, x), y)) -> APP(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
APP(D, app(app(+, x), y)) -> APP(app(+, app(D, x)), app(D, y))


Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

app(D, t) -> 1
app(D, constant) -> 0
app(D, app(app(+, x), y)) -> app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) -> app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) -> app(app(-, app(D, x)), app(D, y))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(t)=  0  
  POL(0)=  0  
  POL(1)=  0  
  POL(constant)=  0  
  POL(*)=  0  
  POL(D)=  1  
  POL(-)=  0  
  POL(app(x1, x2))=  0  
  POL(+)=  0  
  POL(APP(x1, x2))=  x1  

resulting in one new DP problem. 



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


Dependency Pairs:

APP(D, app(app(-, x), y)) -> APP(D, y)
APP(D, app(app(-, x), y)) -> APP(D, x)
APP(D, app(app(*, x), y)) -> APP(D, y)
APP(D, app(app(*, x), y)) -> APP(D, x)
APP(D, app(app(+, x), y)) -> APP(D, y)
APP(D, app(app(+, x), y)) -> APP(D, x)


Rules:


app(D, t) -> 1
app(D, constant) -> 0
app(D, app(app(+, x), y)) -> app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) -> app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) -> app(app(-, app(D, x)), app(D, y))





The following dependency pairs can be strictly oriented:

APP(D, app(app(-, x), y)) -> APP(D, y)
APP(D, app(app(-, x), y)) -> APP(D, x)


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(*)=  0  
  POL(D)=  0  
  POL(-)=  1  
  POL(APP(x1, x2))=  1 + x2  
  POL(app(x1, x2))=  x1 + x2  
  POL(+)=  0  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Polo
             ...
               →DP Problem 3
                 ↳Polynomial Ordering


Dependency Pairs:

APP(D, app(app(*, x), y)) -> APP(D, y)
APP(D, app(app(*, x), y)) -> APP(D, x)
APP(D, app(app(+, x), y)) -> APP(D, y)
APP(D, app(app(+, x), y)) -> APP(D, x)


Rules:


app(D, t) -> 1
app(D, constant) -> 0
app(D, app(app(+, x), y)) -> app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) -> app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) -> app(app(-, app(D, x)), app(D, y))





The following dependency pairs can be strictly oriented:

APP(D, app(app(*, x), y)) -> APP(D, y)
APP(D, app(app(*, x), y)) -> APP(D, x)


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

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

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Polo
             ...
               →DP Problem 4
                 ↳Polynomial Ordering


Dependency Pairs:

APP(D, app(app(+, x), y)) -> APP(D, y)
APP(D, app(app(+, x), y)) -> APP(D, x)


Rules:


app(D, t) -> 1
app(D, constant) -> 0
app(D, app(app(+, x), y)) -> app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) -> app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) -> app(app(-, app(D, x)), app(D, y))





The following dependency pairs can be strictly oriented:

APP(D, app(app(+, x), y)) -> APP(D, y)
APP(D, app(app(+, x), y)) -> APP(D, x)


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

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

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Polo
             ...
               →DP Problem 5
                 ↳Dependency Graph


Dependency Pair:


Rules:


app(D, t) -> 1
app(D, constant) -> 0
app(D, app(app(+, x), y)) -> app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) -> app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) -> app(app(-, app(D, x)), app(D, y))





Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes