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

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains three SCCs. 


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


Dependency Pair:

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


Rules:


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





The following dependency pair can be strictly oriented:

APP(f, app(s, x)) -> APP(f, 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(s)=  0  
  POL(APP(x1, x2))=  x2  
  POL(f)=  0  
  POL(app(x1, x2))=  1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 4
             ↳Dependency Graph
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo


Dependency Pair:


Rules:


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





Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

APP(g, app(app(cons, 0), y)) -> APP(g, y)


Rules:


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





The following dependency pair can be strictly oriented:

APP(g, app(app(cons, 0), y)) -> APP(g, y)


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)=  0  
  POL(g)=  0  
  POL(cons)=  0  
  POL(APP(x1, x2))=  x2  
  POL(app(x1, x2))=  1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 5
             ↳Dependency Graph
       →DP Problem 3
         ↳Polo


Dependency Pair:


Rules:


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





Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

APP(h, app(app(cons, x), y)) -> APP(h, app(g, app(app(cons, x), y)))


Rules:


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





The following dependency pair can be strictly oriented:

APP(h, app(app(cons, x), y)) -> APP(h, app(g, app(app(cons, x), y)))


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

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


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(g)=  0  
  POL(cons)=  0  
  POL(s)=  0  
  POL(h)=  0  
  POL(app(x1, x2))=  1 + x1  
  POL(f)=  0  
  POL(APP(x1, x2))=  1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
           →DP Problem 6
             ↳Dependency Graph


Dependency Pair:


Rules:


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





Using the Dependency Graph resulted in no new DP problems.

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