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

Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains two SCCs.


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


Dependency Pair:

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


Rules:


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





The following dependency pair can be strictly oriented:

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


There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(INT_LIST(x1))=  x1  
  POL(.(x1, x2))=  1 + x2  

resulting in one new DP problem.



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


Dependency Pair:


Rules:


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


Rules:


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





The following dependency pair can be strictly oriented:

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


There are no usable rules using the Ce-refinement 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:


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





Using the Dependency Graph resulted in no new DP problems.

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