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)

Innermost 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
         ↳Usable Rules (Innermost)
       →DP Problem 2
         ↳UsableRules


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)


Strategy:

innermost




As we are in the innermost case, we can delete all 7 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
           →DP Problem 3
             ↳Size-Change Principle
       →DP Problem 2
         ↳UsableRules


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)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. CHOOSE(x, cons(v, w), s(y), s(z)) -> CHOOSE(x, cons(v, w), y, z)
  2. CHOOSE(x, cons(v, w), 0, s(z)) -> INSERT(x, w)
  3. INSERT(x, cons(v, w)) -> CHOOSE(x, cons(v, w), x, v)
and get the following Size-Change Graph(s):
{1} , {1}
1=1
2=2
3>3
4>4
{2} , {2}
1=1
2>2
{3} , {3}
1=1
1=3
2=2
2>4

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1=1
2=2
3>3
4>4
{2} , {3}
1=1
1=3
2>2
2>4
{3} , {2}
1=1
2>2
{1} , {3}
1=1
1=3
2>2
2>4
{2} , {1}
1=1
1>3
2>2
2>4
{1} , {1}
1=1
1>3
2>2
2>4

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
cons(x1, x2) -> cons(x1, x2)
s(x1) -> s(x1)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳Usable Rules (Innermost)


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)


Strategy:

innermost




As we are in the innermost case, we can delete all 7 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
           →DP Problem 4
             ↳Size-Change Principle


Dependency Pair:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. SORT(cons(x, y)) -> SORT(y)
and get the following Size-Change Graph(s):
{1} , {1}
1>1

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1>1

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
cons(x1, x2) -> cons(x1, x2)

We obtain no new DP problems.

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