S1: ∀x ∃y ∀z [ x+ y = z]
For all x, there exist y such that for all z. [x + y = z]. This is false.
Counter example: x = 2 and y = 1 then x + y = 3. Then there is only one value of z that satisfy this. And that is z = 3. But here it is given for all z. So it should be true for z =1 , z = 2, z = 0. Clearly it is not true for all z.
S2: ∃x ∀y ∃z [x + y = z]
There exist x, such that for all y, there exist z which satisfy [x + y = z]. This is true.
Let take any x, x = 2 then for all y, there exist z which satisfy [y = z - 2]. This is true. Take what ever value of y there is always exist a z. which satisfy this. If y = 4 then z = 6. If y = 5 then z = 7.
What if i take x = 3, then also this is true.
So ∀x ∀y ∃z [x + y = z] this is also true.