I have looked all over the web and can't find any elegant proofs for the commutative, associative and distributive laws of Sets:
Commutative Law
$$A\cup B = B\cup A, \ A\cap B = B\cap A$$
Associative Law
$$A\cup (B\cup C) = (A\cup B)\cup C, \ A\cap (B\cap C) = (A\cap B)\cap C$$
Distributive Law
$$A\cap (B\cup C) = (A\cap B)\cup (A\cap C), A\cup (B\cap C) = (A\cup B)\cap (A\cup C) $$
How would these be proved in an elegant way?
The question comes from Tom M Apostol Calculus Volume I book. He gives an example proof of the commutative law as follows.
Let $X=A\cup B$, $Y=B\cup A$. To prove that $X=Y$ we prove that $X\subseteq Y$ and $Y\subseteq X$. Suppose that $x\in X$. Then $x$ is in at least one of $A$ or $B$. Hence, $x$ is in at least one of $B$ or $A$; so $x\in Y$. Thus, every element of $X$ is also in $Y$, so $X\subseteq Y$. Similary, we find that $Y\subseteq X$, so $X=Y$
He doesn't presuppose knowledge of truth tables. Can this proof be written in a more mathematical and concise way? How would I go about writing concise proofs for the other laws?
I understand the logic, but I am new to mathematical language.
Best Answer
You can use truth tables. Let $0$ indicate $x\not\in X$ and $1$ indicate $x\in X$, where $X$ is a set.
For example, the first distributive law:
$$\begin{array} {ccc|c|c|c|c|c} A&B&C&(B\cup C)&A\cap (B\cup C)&(A\cap B)&(A\cap C)&(A\cap B)\cup (A\cap C)\\ \hline 0&0&0&0&0&0&0&0\\ 0&0&1&1&0&0&0&0\\ 0&1&0&1&0&0&0&0\\ 0&1&1&1&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ 1&0&1&1&1&0&1&1\\ 1&1&0&1&1&1&0&1\\ 1&1&1&1&1&1&1&1 \end{array} $$
Column $5$ equals column $8$.
In response to the updated question, look at the definitions of union and intersection:
$$A\cup B = \{x|x\in A \lor x\in B\}$$ $$A\cap B = \{x|x\in A \land x\in B\}$$
Both $\lor$ and $\land$ are commutative for example (but you need a truth table to prove it), and so, by extending this concept over the definitions of $\cup$ and $\cap$, are these operators, i.e.:
$$A\cup B = \{x|x\in A \lor x\in B\} = \{x|x\in B \lor x\in A\} = B\cup A$$