How can I prove that these two sets are equal?

Question

I understand that a two way proof is necessary to show that the two sets have the same members, but I don't know if I have achieved the correct proof. This is what I have written so far (I am very new to set theory):

Show that z\(x V y) = (z\x) ∧ (z\y)

z\(x ∨ y)

   (x ∨ y) = a = {a : a ∈ x or a ∈ y}
   z\(x ∨ y) = μ = {μ ∈ z : μ ∉ a} = {μ ∈ z: μ ∉ x and μ ∉ y}
   {μ ∈ z: μ ∉ x and μ ∉ y} = (z\x) ∧ (z\y)

(z\x) ∧ (z\y)

   (z\x) = b = {b ∈ z : b ∉ x}
   (z\y) = c = {c ∈ z : c ∉ y}
   (z\x) ∧ (z\y) = μ = {μ ∈ z: μ ∉ x and μ ∉ y}
   μ = {μ ∈ z: μ ∉ x and μ ∉ y} = z\(x ∨ y)

z\(x ∨ y) = (z\x) ∧ (z\y)

Does this prove anything? I feel like I have just restated the different sides of the equations as sets and then reconnected them. Any comments would be appreciated.

Answer 1

Hint

Due to the fact that in set theory :

$A=B \leftrightarrow \forall x(x \in A \leftrightarrow x \in B)$

it is correct to say that for "a two way proof is necessary to show that the two sets have the same members".

You can do it step-by-step, "unwinding" the definitions.

Thus :

$x \in Z \setminus (X \cup Y)$ iff $x \in Z \land x \notin (X \cup Y)$.

In turn, this gives us :

$x \in Z \setminus (X \cup Y)$ iff $x \in Z \land (x \notin X \land x \notin Y)$,

and thus, finally :

$x \in Z \setminus (X \cup Y)$ iff $(x \in Z \land x \notin X) \land (x \in Z \land x \notin Y)$.