Term Rewriting System R:
[x, y, z]
norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

NORM(g(x, y)) -> NORM(x)
F(x, g(y, z)) -> F(x, y)
REM(g(x, y), s(z)) -> REM(x, z)

Furthermore, R contains three SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS


Dependency Pair:

NORM(g(x, y)) -> NORM(x)


Rules:


norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)





The following dependency pair can be strictly oriented:

NORM(g(x, y)) -> NORM(x)


There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(g(x1, x2))=  1 + x1 + x2  
  POL(NORM(x1))=  x1  

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


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


Dependency Pair:


Rules:


norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Argument Filtering and Ordering
       →DP Problem 3
         ↳AFS


Dependency Pair:

F(x, g(y, z)) -> F(x, y)


Rules:


norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)





The following dependency pair can be strictly oriented:

F(x, g(y, z)) -> F(x, y)


There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(g(x1, x2))=  1 + x1 + x2  
  POL(F(x1, x2))=  x1 + x2  

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


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


Dependency Pair:


Rules:


norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Argument Filtering and Ordering


Dependency Pair:

REM(g(x, y), s(z)) -> REM(x, z)


Rules:


norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)





The following dependency pair can be strictly oriented:

REM(g(x, y), s(z)) -> REM(x, z)


There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(g(x1, x2))=  1 + x1 + x2  
  POL(REM(x1, x2))=  x1 + x2  
  POL(s(x1))=  x1  

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


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


Dependency Pair:


Rules:


norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)





Using the Dependency Graph resulted in no new DP problems.

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