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 monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        |sort|#_A(x1) = x1
        cons_A(x1,x2) = x1 + x2 + 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        cons > |sort|#
      
      argument filter:
    
        pi(|sort|#) = 1
        pi(cons) = [1, 2]
    

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7

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. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        insert#_A(x1,x2) = x2 + 1
        cons_A(x1,x2) = x1 + x2 + 2
        choose#_A(x1,x2,x3,x4) = x2
        s_A(x1) = 1
        |0|_A() = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        insert# > |0| > choose# > cons > s
      
      argument filter:
    
        pi(insert#) = 2
        pi(cons) = [1, 2]
        pi(choose#) = [2]
        pi(s) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1, p3

We remove them from the problem.

-- SCC decomposition.

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

p1: 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:

  {p1}


-- Reduction pair.

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

p1: 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 set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        choose#_A(x1,x2,x3,x4) = x3 + x4
        cons_A(x1,x2) = x1 + x2
        s_A(x1) = x1 + 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        cons > choose# > s
      
      argument filter:
    
        pi(choose#) = 4
        pi(cons) = [1, 2]
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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