Term Rewriting System R:
[x, y, v, w, z]
sort(nil) -> nil
sort(cons(x, y)) -> insert(x, sort(y))
insert(x, nil) -> cons(x, nil)
insert(x, cons(v, w)) -> choose(x, cons(v, w), x, v)
choose(x, cons(v, w), y, 0) -> cons(x, cons(v, w))
choose(x, cons(v, w), 0, s(z)) -> cons(v, insert(x, w))
choose(x, cons(v, w), s(y), s(z)) -> choose(x, cons(v, w), y, z)

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

SORT(cons(x, y)) -> INSERT(x, sort(y))
SORT(cons(x, y)) -> SORT(y)
INSERT(x, cons(v, w)) -> CHOOSE(x, cons(v, w), x, v)
CHOOSE(x, cons(v, w), 0, s(z)) -> INSERT(x, w)
CHOOSE(x, cons(v, w), s(y), s(z)) -> CHOOSE(x, cons(v, w), y, z)

Furthermore, R contains two SCCs. 


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


Dependency Pairs:

CHOOSE(x, cons(v, w), s(y), s(z)) -> CHOOSE(x, cons(v, w), y, z)
CHOOSE(x, cons(v, w), 0, s(z)) -> INSERT(x, w)
INSERT(x, cons(v, w)) -> CHOOSE(x, cons(v, w), x, v)


Rules:


sort(nil) -> nil
sort(cons(x, y)) -> insert(x, sort(y))
insert(x, nil) -> cons(x, nil)
insert(x, cons(v, w)) -> choose(x, cons(v, w), x, v)
choose(x, cons(v, w), y, 0) -> cons(x, cons(v, w))
choose(x, cons(v, w), 0, s(z)) -> cons(v, insert(x, w))
choose(x, cons(v, w), s(y), s(z)) -> choose(x, cons(v, w), y, z)





The following dependency pair can be strictly oriented:

CHOOSE(x, cons(v, w), 0, s(z)) -> INSERT(x, w)


The following rules can be oriented:

sort(nil) -> nil
sort(cons(x, y)) -> insert(x, sort(y))
insert(x, nil) -> cons(x, nil)
insert(x, cons(v, w)) -> choose(x, cons(v, w), x, v)
choose(x, cons(v, w), y, 0) -> cons(x, cons(v, w))
choose(x, cons(v, w), 0, s(z)) -> cons(v, insert(x, w))
choose(x, cons(v, w), s(y), s(z)) -> choose(x, cons(v, w), y, z)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(cons(x1, x2))=  1 + x1 + x2  
  POL(nil)=  0  
  POL(sort(x1))=  x1  
  POL(insert(x1, x2))=  1 + x1 + x2  
  POL(choose(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem.
Used Argument Filtering System:
CHOOSE(x1, x2, x3, x4) -> x2
cons(x1, x2) -> cons(x1, x2)
INSERT(x1, x2) -> x2
sort(x1) -> sort(x1)
insert(x1, x2) -> insert(x1, x2)
choose(x1, x2, x3, x4) -> choose(x1, x2)


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


Dependency Pairs:

CHOOSE(x, cons(v, w), s(y), s(z)) -> CHOOSE(x, cons(v, w), y, z)
INSERT(x, cons(v, w)) -> CHOOSE(x, cons(v, w), x, v)


Rules:


sort(nil) -> nil
sort(cons(x, y)) -> insert(x, sort(y))
insert(x, nil) -> cons(x, nil)
insert(x, cons(v, w)) -> choose(x, cons(v, w), x, v)
choose(x, cons(v, w), y, 0) -> cons(x, cons(v, w))
choose(x, cons(v, w), 0, s(z)) -> cons(v, insert(x, w))
choose(x, cons(v, w), s(y), s(z)) -> choose(x, cons(v, w), y, z)





Using the Dependency Graph the DP problem was split into 1 DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 3
             ↳DGraph
             ...
               →DP Problem 4
                 ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳AFS


Dependency Pair:

CHOOSE(x, cons(v, w), s(y), s(z)) -> CHOOSE(x, cons(v, w), y, z)


Rules:


sort(nil) -> nil
sort(cons(x, y)) -> insert(x, sort(y))
insert(x, nil) -> cons(x, nil)
insert(x, cons(v, w)) -> choose(x, cons(v, w), x, v)
choose(x, cons(v, w), y, 0) -> cons(x, cons(v, w))
choose(x, cons(v, w), 0, s(z)) -> cons(v, insert(x, w))
choose(x, cons(v, w), s(y), s(z)) -> choose(x, cons(v, w), y, z)





The following dependency pair can be strictly oriented:

CHOOSE(x, cons(v, w), s(y), s(z)) -> CHOOSE(x, cons(v, w), y, z)


The following rules can be oriented:

sort(nil) -> nil
sort(cons(x, y)) -> insert(x, sort(y))
insert(x, nil) -> cons(x, nil)
insert(x, cons(v, w)) -> choose(x, cons(v, w), x, v)
choose(x, cons(v, w), y, 0) -> cons(x, cons(v, w))
choose(x, cons(v, w), 0, s(z)) -> cons(v, insert(x, w))
choose(x, cons(v, w), s(y), s(z)) -> choose(x, cons(v, w), y, z)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(CHOOSE(x1, x2, x3, x4))=  x1 + x2 + x3 + x4  
  POL(cons(x1, x2))=  x1 + x2  
  POL(nil)=  0  
  POL(s(x1))=  1 + x1  
  POL(sort(x1))=  x1  
  POL(insert(x1, x2))=  x1 + x2  
  POL(choose(x1, x2))=  x1 + x2  

resulting in one new DP problem.
Used Argument Filtering System:
CHOOSE(x1, x2, x3, x4) -> CHOOSE(x1, x2, x3, x4)
cons(x1, x2) -> cons(x1, x2)
s(x1) -> s(x1)
sort(x1) -> sort(x1)
insert(x1, x2) -> insert(x1, x2)
choose(x1, x2, x3, x4) -> choose(x1, x2)


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


Dependency Pair:


Rules:


sort(nil) -> nil
sort(cons(x, y)) -> insert(x, sort(y))
insert(x, nil) -> cons(x, nil)
insert(x, cons(v, w)) -> choose(x, cons(v, w), x, v)
choose(x, cons(v, w), y, 0) -> cons(x, cons(v, w))
choose(x, cons(v, w), 0, s(z)) -> cons(v, insert(x, w))
choose(x, cons(v, w), s(y), s(z)) -> choose(x, cons(v, w), y, z)





Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

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


Rules:


sort(nil) -> nil
sort(cons(x, y)) -> insert(x, sort(y))
insert(x, nil) -> cons(x, nil)
insert(x, cons(v, w)) -> choose(x, cons(v, w), x, v)
choose(x, cons(v, w), y, 0) -> cons(x, cons(v, w))
choose(x, cons(v, w), 0, s(z)) -> cons(v, insert(x, w))
choose(x, cons(v, w), s(y), s(z)) -> choose(x, cons(v, w), y, z)





The following dependency pair can be strictly oriented:

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


The following rules can be oriented:

sort(nil) -> nil
sort(cons(x, y)) -> insert(x, sort(y))
insert(x, nil) -> cons(x, nil)
insert(x, cons(v, w)) -> choose(x, cons(v, w), x, v)
choose(x, cons(v, w), y, 0) -> cons(x, cons(v, w))
choose(x, cons(v, w), 0, s(z)) -> cons(v, insert(x, w))
choose(x, cons(v, w), s(y), s(z)) -> choose(x, cons(v, w), y, z)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(cons(x1, x2))=  1 + x1 + x2  
  POL(nil)=  0  
  POL(sort(x1))=  x1  
  POL(insert(x1, x2))=  1 + x1 + x2  
  POL(choose(x1, x2))=  1 + x1 + x2  
  POL(SORT(x1))=  1 + x1  

resulting in one new DP problem.
Used Argument Filtering System:
SORT(x1) -> SORT(x1)
cons(x1, x2) -> cons(x1, x2)
sort(x1) -> sort(x1)
insert(x1, x2) -> insert(x1, x2)
choose(x1, x2, x3, x4) -> choose(x1, x2)


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


Dependency Pair:


Rules:


sort(nil) -> nil
sort(cons(x, y)) -> insert(x, sort(y))
insert(x, nil) -> cons(x, nil)
insert(x, cons(v, w)) -> choose(x, cons(v, w), x, v)
choose(x, cons(v, w), y, 0) -> cons(x, cons(v, w))
choose(x, cons(v, w), 0, s(z)) -> cons(v, insert(x, w))
choose(x, cons(v, w), s(y), s(z)) -> choose(x, cons(v, w), y, z)





Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:01 minutes