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))

Innermost 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
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳AFS


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))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
INTLIST(x1) -> INTLIST(x1)
.(x1, x2) -> .(x1, x2)


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


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))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Argument Filtering and 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))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
INT(x1, x2) -> x2
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
           →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))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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