YES

We show the termination of the TRS R:

  sum(|0|()) -> |0|()
  sum(s(x)) -> +(sqr(s(x)),sum(x))
  sqr(x) -> *(x,x)
  sum(s(x)) -> +(*(s(x),s(x)),sum(x))

-- SCC decomposition.

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

p1: sum#(s(x)) -> sqr#(s(x))
p2: sum#(s(x)) -> sum#(x)
p3: sum#(s(x)) -> sum#(x)

and R consists of:

r1: sum(|0|()) -> |0|()
r2: sum(s(x)) -> +(sqr(s(x)),sum(x))
r3: sqr(x) -> *(x,x)
r4: sum(s(x)) -> +(*(s(x),s(x)),sum(x))

The estimated dependency graph contains the following SCCs:

  {p2, p3}


-- Reduction pair.

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

p1: sum#(s(x)) -> sum#(x)

and R consists of:

r1: sum(|0|()) -> |0|()
r2: sum(s(x)) -> +(sqr(s(x)),sum(x))
r3: sqr(x) -> *(x,x)
r4: sum(s(x)) -> +(*(s(x),s(x)),sum(x))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        s > sum#
      
      argument filter:
    
        pi(sum#) = [1]
        pi(s) = [1]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > sum#
      
      argument filter:
    
        pi(sum#) = [1]
        pi(s) = [1]
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        s > sum#
      
      argument filter:
    
        pi(sum#) = [1]
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4

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