Proof of intersection and union of Set A with Empty Set I need to prove the following:

Prove that $A\cup \!\, \varnothing \!\,=A$ and $A\cap \!\, \varnothing \!\,=\varnothing \!\,$

It's my understanding that to prove equality, I must prove that both are subsets of each other.
So to prove $A\cup \!\, \varnothing \!\,=A$, we need to prove that $A\cup \!\, \varnothing \!\,\subseteq \!\,A$ and $A\subseteq \!\,A\cup \!\, \varnothing \!\,$.

However, I found an example proof for $A \cup \!\, A$ in my book and I adapted it and got this:

$A\cup \!\, \varnothing \!\,=$ {$x:x\in \!\, A  \ \text{or} \ x\in \!\, \varnothing \!\,$}
= {$x:x\in \!\, A$} = A

$A\cap \!\,  \varnothing \!\,=$ {$x:x\in \!\, A  \ \text{and} \ x\in \!\, \varnothing \!\,$}
= {$x:x\in \!\, \varnothing \!\,$} = $\varnothing \!\,$

Do my proofs look ok?
This looks fine, but you could point out a few more details. For instance, $x\in \varnothing$ is always false. Therefore
$x \in A \text{ or } x\in \varnothing
$
is logically equivalent to
$
x \in A
$
and therefore the two set descriptions
$$
\{x \mid x \in A \text{ or } x \in \varnothing\},\quad \{x\mid x \in A\} 
$$
must describe the same set, since the conditions are true for exactly the same elements $x$.

Similarily, because $x \in \varnothing$ is trivially false, the condition $x \in A \text{ and } x \in \varnothing$ will always be false, so the two set descriptions
$$
\{x \mid x \in A \text{ and } x \in \varnothing\},\quad \{x\mid x \in \varnothing \} 
$$
must describe the same set.

Of course, for any set $B$ we have
$$
B = \{x \mid x \in B\}
$$
we need to proof that A U phi=A,
it can be written as,
A U PHI={X:X e A OR X e phi}
to do it in a simpleast way I will use a example,
write in roaster form
A={1,2,3}
PHI={4,2,5}
WHEN YOU WRITE THE UNION IT COMES OUT TO BE {1,2,3,4,5}
THEREFORE AUPHI=A