Exercise 1.4.1

Proof. For all $x$, we have that

which shows that $A\cap B=B\cap A$. Similarly, we have that

which proves that $A\cup B=B\cup A$. □

Proof. Again, for any $x$, we have that

showing that $(A\cap B)\cap C=A\cap (B\cap C)$. Likewise, we have

and hence $(A\cup B)\cup C=A\cup (B\cup C)$. □

Proof. For all $x$, we have

which proves the result. □

Proof. For every $x$, we have that

proving the desired result. □

Proof. For all $x$,

of course showing the result. □

Proof. Again, for all $x$, we have

which proves the result. □

Proof. For all $x$, we have

showing the result. □

Proof. $\text{(→)}$ First suppose that $A-B=\varnothing$ and consider any $x\in A$. Suppose that $x\notin B$. Then we would have $x\in A$ but $x\notin B$ so that $x\in A-B$, contradicting the fact that $A-B=\varnothing$. Hence, it has to be that $x\in B$ so that $A\subseteq B$ since $x$ was arbitrary.

$\text{(←)}$ Now suppose that $A\subseteq B$ but that $A-B\ne \varnothing$ so that there is an $x\in A-B$. Therefore, $x\in A$ but $x\notin B$. However, since $A\subseteq B$ it must also be that $x\in B$ since $x\in A$, a contradiction! So it in fact has to be that $A-B=\varnothing$ as desired. □

Proof. This is very nearly self-evident, but suppose $A-A\ne \varnothing$ so that there is an $x\in A-A$. Then it would both be true that $x\in A$ and $x\notin A$. As this is a clear contradiction, it has to be that in fact $A-A=\varnothing$ as claimed. □

Proof. By definition and the above lemma,

Clearly $\varnothing \cup \varnothing =\varnothing$ since, were there an $x\in \varnothing \cup \varnothing$, we would have $x\in \varnothing$ or $x\in \varnothing$, the same impossibility in either case. □

Proof. This follows trivially from the commutativity of the union operation, proven above. By definition, we simply have

□

Proof. To prove this, we work out the truth table for the right-hand side, in which we set $R=(P\wedge \neg <!--\; FUNCTION\; APPLICATION\; -->Q)\vee (Q\wedge \neg <!--\; FUNCTION\; APPLICATION\; -->P)$ for brevity:

Since the truth table for $R$ is identical to that for $P⊻Q$ above, this proves the identity. □

Proof. This is again most easily proved via exhaustive truth tables:

Since the final column of both tables are identical, this shows the desired result. □

Proof. For any $x$, define the proposition $P$ as $x\in A$ and $Q$ as $x\in B$. Then we have that

which of course proves the result. □

Proof. For any $x$, define the proposition $P$ as $x\in A$, $Q$ as $x\in B$ and $R$ as $x\in C$. We then have that

This of course suffices to show that $(A△B)△C=A△(B△C)$. □