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^1
    order: standard order
    interpretations:
      |sort|#_A(x1) = x1
      cons_A(x1,x2) = x2 + 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^1
    order: standard order
    interpretations:
      insert#_A(x1,x2) = x2
      cons_A(x1,x2) = x2 + 1
      choose#_A(x1,x2,x3,x4) = x2
      s_A(x1) = x1 + 1
      |0|_A() = 1

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

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)

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}


-- 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:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      choose#_A(x1,x2,x3,x4) = x3 + x4
      cons_A(x1,x2) = 0
      s_A(x1) = x1 + 1

The next rules are strictly ordered:

  p1

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