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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      |sort|#_A(x1) = ((1,0,0),(0,0,0),(1,1,0)) x1
      cons_A(x1,x2) = ((1,0,0),(1,1,0),(1,1,1)) x1 + ((1,0,0),(1,1,0),(1,1,1)) x2 + (1,1,1)

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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      insert#_A(x1,x2) = x2 + (1,0,1)
      cons_A(x1,x2) = ((0,0,0),(0,0,0),(1,0,0)) x1 + ((1,0,0),(0,0,0),(1,1,0)) x2 + (1,1,0)
      choose#_A(x1,x2,x3,x4) = x2 + ((0,0,0),(1,0,0),(0,1,0)) x3
      s_A(x1) = ((1,0,0),(1,1,0),(1,1,0)) x1 + (1,1,1)
      |0|_A() = (0,2,1)

The next rules are strictly ordered:

  p1, p2, p3

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