Currently studying the first chapter of Munkre's Topology, and I'm enjoying time spent proving statements. I am trying to prove De Morgan's laws. I am worried that this proof is not valid, specifically in the logical jumps such as "If $a\notin (Y\cup Z)$ then $a\notin Y$ and $a\notin Z$." To obtain that implication I used combinations of $a\in Y$, $a\in Z$, $a\notin Y$ or $a\notin Z$ and got
$$a\in Y\lor a\in Z \Rightarrow a\in (Y \cup Z)$$
$$a\notin Y\land a\notin Z \Rightarrow a\notin (Y\cup Z)$$
Then an part of this question is: Should that process be outlined in the proof?
Overall: Is the following proof valid? This proof is left as an exercise in the main text body. I've wiped out some fallacies I was using but I'm still insecure about the proof.
Theorem. Let $X,Y,Z$ be sets. Then $X \setminus (Y\cup Z)=(X\setminus Y)\cap (X \setminus Z)$.
Proof. If $a\in X\setminus (Y\cup Z)$ then $a\in X$ and $a\notin (Y\cup Z)$. If $a\notin (Y\cup Z)$ then $a\notin Y$ and $a\notin Z$. Then if $a\in X\setminus (Y\cup Z)$ then $a\in X$ and $a\notin Y$ and $a\notin Z$.
If $a\in (X\setminus Y)\cap (X \setminus Z)$ then $a\in (X\setminus Y)$ and $a\in (X\setminus Z)$. If $a\in (X\setminus Y)$ then $a\in X$ and $a\notin Y$. If $a\in (X\setminus Z)$ then $a\in X$ and $a\notin Z$. Then if $a\in (X\setminus Y)\cap (X \setminus Z)$ then $a\in X$ and $a\notin Y$ and $a\notin Z$.
Then if $a\in X\setminus (Y\cup Z)$ then $a\in (X\setminus Y)\cap (X \setminus Z)$ and if $a\in (X\setminus Y)\cap (X \setminus Z)$ then $a\in X\setminus (Y\cup Z)$. Then $a\in X\setminus (Y\cup Z)$ iff $a\in (X\setminus Y)\cap (X \setminus Z)$. Then, using the axiom of extension, $X \setminus (Y\cup Z)=(X\setminus Y)\cap (X \setminus Z)$. $\square$
Best Answer
You are on the right track! Well done!
However, if I were you, I would stress out that $a\in X$ and $a\not\in Y$ and $a\not\in Z$ implies $a\in X\backslash Y$ and $a\in X\backslash Z$. Similar observation applies to the reverse relation, that is to say, $a\in X$ and $a\not\in Y$ and $a\not\in Z$ implies that $a\in X\backslash(Y\cup Z)$.
Here is another way to approach it: \begin{align*} X\backslash(Y\cup Z) & = X\cap(Y\cup Z)^{c} \tag{Definition of Difference}\\\\ & = X\cap (Y^{c}\cap Z^{c}) \tag{DeMorgan's Law}\\\\ & = (X\cap X)\cap(Y^{c}\cap Z^{c}) \tag{Idempotence}\\\\ & = X\cap (X\cap(Y^{c}\cap Z^{c})) \tag{Associativity}\\\\ & = X\cap ((X\cap Y^{c})\cap Z^{c}) \tag{Associativity}\\\\\ & = (X\cap(X\cap Y^{c}))\cap Z^{c} \tag{Associativity}\\\\ & = ((X\cap Y^{c})\cap X)\cap Z^{c} \tag{Commutativity}\\\\ & = (X\cap Y^{c})\cap (X\cap Z^{c}) \tag{Associativity}\\\\ & = (X\backslash Y)\cap(X\backslash Z) \tag{Definition of Difference} \end{align*}
Hopefully this helps!