Term Rewriting System R:
[x, y]
intlist(nil) -> nil
intlist(cons(x, y)) -> cons(s(x), intlist(y))
intlist(cons(x, nil)) -> cons(s(x), nil)
int(s(x), 0) -> nil
int(x, x) -> cons(x, nil)
int(s(x), s(y)) -> intlist(int(x, y))
int(0, s(y)) -> cons(0, int(s(0), s(y)))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

INTLIST(cons(x, y)) -> INTLIST(y)
INT(s(x), s(y)) -> INTLIST(int(x, y))
INT(s(x), s(y)) -> INT(x, y)
INT(0, s(y)) -> INT(s(0), s(y))

Furthermore, R contains two SCCs. 


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


Dependency Pair:

INTLIST(cons(x, y)) -> INTLIST(y)


Rules:


intlist(nil) -> nil
intlist(cons(x, y)) -> cons(s(x), intlist(y))
intlist(cons(x, nil)) -> cons(s(x), nil)
int(s(x), 0) -> nil
int(x, x) -> cons(x, nil)
int(s(x), s(y)) -> intlist(int(x, y))
int(0, s(y)) -> cons(0, int(s(0), s(y)))





The following dependency pair can be strictly oriented:

INTLIST(cons(x, y)) -> INTLIST(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(cons(x1, x2))=  1 + x2  
  POL(INTLIST(x1))=  x1  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


intlist(nil) -> nil
intlist(cons(x, y)) -> cons(s(x), intlist(y))
intlist(cons(x, nil)) -> cons(s(x), nil)
int(s(x), 0) -> nil
int(x, x) -> cons(x, nil)
int(s(x), s(y)) -> intlist(int(x, y))
int(0, s(y)) -> cons(0, int(s(0), s(y)))





Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pairs:

INT(0, s(y)) -> INT(s(0), s(y))
INT(s(x), s(y)) -> INT(x, y)


Rules:


intlist(nil) -> nil
intlist(cons(x, y)) -> cons(s(x), intlist(y))
intlist(cons(x, nil)) -> cons(s(x), nil)
int(s(x), 0) -> nil
int(x, x) -> cons(x, nil)
int(s(x), s(y)) -> intlist(int(x, y))
int(0, s(y)) -> cons(0, int(s(0), s(y)))





The following dependency pair can be strictly oriented:

INT(s(x), s(y)) -> INT(x, 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(s(x1))=  1 + x1  
  POL(INT(x1, x2))=  x2  

resulting in one new DP problem. 



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


Dependency Pair:

INT(0, s(y)) -> INT(s(0), s(y))


Rules:


intlist(nil) -> nil
intlist(cons(x, y)) -> cons(s(x), intlist(y))
intlist(cons(x, nil)) -> cons(s(x), nil)
int(s(x), 0) -> nil
int(x, x) -> cons(x, nil)
int(s(x), s(y)) -> intlist(int(x, y))
int(0, s(y)) -> cons(0, int(s(0), s(y)))





Using the Dependency Graph resulted in no new DP problems.

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