Let $S$ a subspace and $V$ a vector space. Show that the additive identity of $S$ is the additive identity of $V$. Working on the book: Robert Messer. "Linear algebra - The gateway to mathematics" (p. 55)

16. Suppose $S$ is the subspace of a vector space $V$.
a. Show that the additive identity of $S$ is the additive identity of $V$.

This is my attempt to prove it:

*

*$0$ is the additive identity of $V$ and $0'$ is the additive identity of $S$.

*

*Assume $v \in S$

*$v + 0' = v$ ($0'$ is the additive identity of S)

*$v + 0 = v$ ($0$ is the additive identity of V)

*$\forall x(x \in V \to \exists! y(x+y=x))$ (0 is the unique additive identity of $V$)

*$v \in V \to \exists! y(v+y=v)$

*$v \in V$ (since S \subseteq V)

*$\exists! y(v+y=v)$

*

*$v+z=v \land \forall z'(z' \in V \land v+z'=z \to z'=z)$

*$\forall z'(z' \in V \land v+z'=z \to z'=z)$

*$0 \in V \land v+0=v \to 0=z$

*$0' \in V \land v+0'=v \to 0'=z$

*$0=0'$ (by transitivity)

*$\vdots$
Is my proof skeleton correct ?
Would it suffice to show

*

*If $0$ is the additive identity of $V$ then $0'$ is the additive identity of $S$ and,

*If $0'$ is the additive identity of $V$ then $0$ is the additive identity of $S$ ?

Would appreciate some insight from a logic point of view.

My spiel is not too heavy on the logic, but I think it conveys the essential ideas:
Consider:
$\forall v \in V, \; 0_V + v = v + 0_V = v; \tag 1$
now let
$s \in S \subset V; \tag 2$
then
$s \in V, \tag 3$
whence, via (1),
$0_V + s = s; \tag 4$
also, by (2),
$0_S + s = s; \tag 5$
combining (4) and (5) we find
$0_V + s = 0_S + s; \tag 6$
we may now write
$(0_V + s) + (-s) = (0_S + s) + (-s), \tag 7$
from which
$0_V + (s + (-s)) = 0_S + (s + (-s)); \tag 8$
now,
$s + (-s) = 0_V, \tag 9$
so (8) becomes
$0_V + 0_V = 0_S + 0_V, \tag{10}$
and thus by virtue of (1),
$0_V = 0_V + 0_V = 0_S + 0_V = 0_S. \tag{11}$
