I would like the following proof to be reviewed for correctness and clarity of the argument as well as simplicity and wording, in one sentence: Where can I do better?
Proof that $(A \cup B)^c = A^c \cap B^c$.
We will prove the above in two parts, showing that each of the sets $(A \cup B)^c$ and $A^c \cap B^c$ is subset of the other one.
Part 1: Proof that $(A \cup B)^c \subseteq A^c \cap B^c$:
1: Let $x \in (A \cup B)^c$. Then $x \notin A \cup B$.
2: Because $ A \cup B = \{y | y \in A \vee y \in B\}$, it must be that $x \notin A$ and $x \notin B$.
3: As $x \notin A$, it must be that $x \in A^c$, and similarly $x \notin B$ means that $x \in B^c$, we can note that $x \in A^c \cap B^c$.
4: Thus $\forall x(x \in (A \cup B)^c \implies x \in A^c \cap B^c)$, which is $(A \cup B)^c \subseteq A^c \cap B^c$.
Part 2: Proof that $A^c \cap B^c \subseteq (A \cup B)^c$:
1: Let $x \in A^c \cap B^c$, but for contradiction assume $x \notin (A \cup B)^c$.
2: $A \cup B = \{y | y \in A$ or $y \in B\}$, so $x \in A$ or $x \in B$.
3: If $x \in A$, then $x \notin A^c$ and also $x \notin A^c \cap B^c$, similarly if $x \in B$, then $x \notin B^c$ and also $x \notin A^c \cap B^c$.
4: As this means that $x \notin A^c \cap B^c$ we have a contradiction to our initial statement that $x \in A^c \cap B^c$. Therefore the assumption that $x \notin (A \cup B)^c$ must not be the case, meaning that $x \in (A \cup B)^c$.
5: Hence $\forall x(x \in A^c \cap B^c \implies x \in (A \cup B)^c)$, that is $A^c \cap B^c \subseteq (A \cup B)^c.$
Best Answer
As others have pointed out, the proof is correct. But since you are also interested in simplicity, I think it is possible to make it a little less verbose.
For example, part 1 (2) you write
$$ A \cup B = \{y | y \in A \vee y \in B\}$$
I believe that it is safe to assume that anyone who reads your proof knows what $A \cup B$ means. So more concisely, you could just write "Since $x \not \in A \cup B$, we have $x \not \in A$ and $x \not \in B$".
In the part 1 (step 4) you write
$$\forall x(x \in (A \cup B)^c \implies x \in A^c \cap B^c)$$
As my current understanding stands, it is not preferred to use first order logic in proofs. Plain english would be a better choice.
So in the end, you first part could be reduced to:
"Take $x \in (A \cup B)^c$. Then we have $x \not \in A \cup B$, which means that $x \not \in A$ and $x \not \in B$, but then $x \in A^{c}$ and $x \in B^{c}$, and thus $x \in A^{c} \cap B^{c}$. Since $x$ was arbitrary, we have $(A \cup B)^c \subseteq A^c \cap B^c$ ."
Now to the second part:
As in the part 1, I believe stating the definition of $A \cup B$ is not necessary.
And again, if your audience is familiar with a proof by contradiction (most likely they are), in the step 4 you just write "but then $x \not \in A^{c} \cap B^{c}$, a contradiction. Hence if $x \in A^{c} \cap B^{c}$, we must have $x \in (A \cup B)^c$."
And yet again, like in the part 1, I think using plain English instead of the first order logic is a better choice. So instead in the step 5, you could just write "$x$ was arbitrary, hence $A^{c} \cap B^{c} \subseteq (A \cup B)^c$."
So the second part might look like:
"By contradiction. Suppose for some $x$, we have $x \in A^c \cap B^c$ but $x \not \in (A \cup B)^c$. Then $x \in A \cup B$, which means that either $x \in A$ or $x \in B$. If $x \in A$, we have $x \not \in A^{c}$. If $x \in B$, then $x \not \in B^{c}$. In both cases, we have $x \not \in A^{c} \cap B^{c}$, a contradiction. Hence if $x \in A^{c} \cap B^{c}$, we must have $x \in (A \cup B)^c$. Since $x$ was arbitrary, $A^{c} \cap B^{c} \subseteq (A \cup B)^c$."