Which of the following logic statements are true

Question

Question

$S_1: \,\,\forall x \,\exists y \, \forall z\,(x+y=z)$

$S_2: \,\, \exists x \, \forall y \, \exists z\,(x+y=z)$

where $x,y, \text{and} \,\,z $ are real numbers. Which of the following statements are true?

My Approach

I think both $S_1$ and $S_2$ are true. Because $S_1$ says for all $x$,there exists a $y$ whose sum will be $z$, a real number as sum of two real numbers are real.hence $S_1$ holds

same reasoning can be given for $S_2$ .but the answer says $S_2$ only.I know i am making some mistake ,don't know where.please help

Answer 1

$S_1$ is false. Its negation is $$\exists x\forall y\exists z\ x+y\ne z$$ and this is trivially true: take $x=0$, then for any $y$ take $z=y+1$.

$S_2$ is true, since given $x=0$ and some $y$ we can take $z=y$ to make the statement true.

Answer 2

Because $S_1$ says for all $x$, there exists a $y$ whose sum will be $z$, a real number as sum of two real numbers are real.

No, $\forall x \,\exists y \, \forall z\,(x+y=z)$ sais that for any $x$, there is a $y$, such that every $z$ will equal their sum ($x+y$).

Whereas $\exists x \, \forall y \, \exists z\,(x+y=z)$ sais there is an $x$, such that for every $y$, there is a $z$ equal to their sum ($x+y$).