YES

We show the termination of the TRS R:

  |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)

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: |sort|#(cons(x,y)) -> insert#(x,|sort|(y))
p2: |sort|#(cons(x,y)) -> |sort|#(y)
p3: insert#(x,cons(v,w)) -> choose#(x,cons(v,w),x,v)
p4: choose#(x,cons(v,w),|0|(),s(z)) -> insert#(x,w)
p5: choose#(x,cons(v,w),s(y),s(z)) -> choose#(x,cons(v,w),y,z)

and R consists of:

r1: |sort|(nil()) -> nil()
r2: |sort|(cons(x,y)) -> insert(x,|sort|(y))
r3: insert(x,nil()) -> cons(x,nil())
r4: insert(x,cons(v,w)) -> choose(x,cons(v,w),x,v)
r5: choose(x,cons(v,w),y,|0|()) -> cons(x,cons(v,w))
r6: choose(x,cons(v,w),|0|(),s(z)) -> cons(v,insert(x,w))
r7: choose(x,cons(v,w),s(y),s(z)) -> choose(x,cons(v,w),y,z)

The estimated dependency graph contains the following SCCs:

  {p2}
  {p3, p4, p5}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: |sort|#(cons(x,y)) -> |sort|#(y)

and R consists of:

r1: |sort|(nil()) -> nil()
r2: |sort|(cons(x,y)) -> insert(x,|sort|(y))
r3: insert(x,nil()) -> cons(x,nil())
r4: insert(x,cons(v,w)) -> choose(x,cons(v,w),x,v)
r5: choose(x,cons(v,w),y,|0|()) -> cons(x,cons(v,w))
r6: choose(x,cons(v,w),|0|(),s(z)) -> cons(v,insert(x,w))
r7: choose(x,cons(v,w),s(y),s(z)) -> choose(x,cons(v,w),y,z)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          |sort|#_A(x1) = max{3, x1 + 2}
          cons_A(x1,x2) = max{1, x2}
    
      precedence:
      
        |sort|# = cons
      
      partial status:
    
        pi(|sort|#) = [1]
        pi(cons) = [2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          |sort|#_A(x1) = max{3, x1 + 2}
          cons_A(x1,x2) = x2 + 1
    
      precedence:
      
        |sort|# = cons
      
      partial status:
    
        pi(|sort|#) = []
        pi(cons) = [2]
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: insert#(x,cons(v,w)) -> choose#(x,cons(v,w),x,v)
p2: choose#(x,cons(v,w),s(y),s(z)) -> choose#(x,cons(v,w),y,z)
p3: choose#(x,cons(v,w),|0|(),s(z)) -> insert#(x,w)

and R consists of:

r1: |sort|(nil()) -> nil()
r2: |sort|(cons(x,y)) -> insert(x,|sort|(y))
r3: insert(x,nil()) -> cons(x,nil())
r4: insert(x,cons(v,w)) -> choose(x,cons(v,w),x,v)
r5: choose(x,cons(v,w),y,|0|()) -> cons(x,cons(v,w))
r6: choose(x,cons(v,w),|0|(),s(z)) -> cons(v,insert(x,w))
r7: choose(x,cons(v,w),s(y),s(z)) -> choose(x,cons(v,w),y,z)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          insert#_A(x1,x2) = max{10, x2 - 1}
          cons_A(x1,x2) = max{x1 + 14, x2 + 16}
          choose#_A(x1,x2,x3,x4) = max{15, x2 - 2, x4 + 1}
          s_A(x1) = max{8, x1 + 7}
          |0|_A = 0
    
      precedence:
      
        insert# > cons > choose# = |0| > s
      
      partial status:
    
        pi(insert#) = []
        pi(cons) = []
        pi(choose#) = [4]
        pi(s) = [1]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          insert#_A(x1,x2) = 4
          cons_A(x1,x2) = 2
          choose#_A(x1,x2,x3,x4) = 3
          s_A(x1) = x1
          |0|_A = 4
    
      precedence:
      
        insert# > cons = choose# = s = |0|
      
      partial status:
    
        pi(insert#) = []
        pi(cons) = []
        pi(choose#) = []
        pi(s) = [1]
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1, p2, p3

We remove them from the problem.  Then no dependency pair remains.