Suppose $A$ and $B$ are sets. Prove that $\forall x\Bigr(x\in A\Delta B\ \text{iff}\ (x\in A\ \text{iff}\ x\notin B)\Bigr)$.

Question

This is exercise $3.5.18$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:

Suppose $A$ and $B$ are sets. Prove that $\forall x\Bigr(x\in A\Delta B\ \text{iff}\ (x\in A\ \text{iff}\ x\notin B)\Bigr)$.

Here is my proof:

$(\rightarrow)$ Let $x$ be an arbitrary element of $A\Delta B$. Then by definition $x\in(A\setminus B)\cup(B\setminus A)$.

$\quad$$(\rightarrow)$ Now we consider two cases.

Case $1.$ Suppose $x\in A$ and $x\in A\setminus B$. So $x\notin B$. Ergo if $x\in A$ then $x\notin B$.

Case $2.$ Suppose $x\in B$ and $x\in B\setminus A$. So $x\notin A$. Ergo if $x\in B$ then $x\notin A$. Therefore if $x\in A$ then $x\notin B$.

Since the above cases are exhaustive, if $x\in A$ then $x\notin B$. Thus if $x\in A\Delta B$ then if $x\in A$ then $x\notin B$.

$\quad$$(\leftarrow)$ Now we consider two cases.

Case $1.$ Suppose $x\notin B$ and $x\in A\setminus B$. So $x\in A$. Ergo if $x\notin B$ then $x\in A$.

Case $2.$ Suppose $x\notin A$ and $x\in B\setminus A$. So $x\in B$. Ergo if $x\notin A$ then $x\in B$. Therefore if $x\notin B$ then $x\in A$.

Since the above cases are exhaustive, if $x\notin B$ then $x\in A$. Thus if $x\in A\Delta B$ then if $x\notin B$ then $x\in A$.

Since $x$ is arbitrary, $\forall x\Bigr(x\in A\Delta B\rightarrow(x\in A\ \text{iff}\ x\notin B)\Bigr).$

$(\leftarrow)$ Let $x$ be arbitrary such that $x\in A$ iff $x\notin B$. Now we consider two cases.

Case $1.$ Suppose $x\in A\setminus B$. Therefore $x\in(A\setminus B)\cup(B\setminus A)$ and so $x\in A\Delta B$.

Case $2.$ Suppose $x\notin (A\setminus B)$. This means $x\notin A$ or $x\in B$. Now we consider two cases.

Case $2.1.$ Suppose $x\notin A$. Since $x\in A$ iff $x\notin B$, $x\in B$. Ergo $x\in B\setminus A$.

Case $2.2.$ Suppose $x\in B$. Since $x\in A$ iff $x\notin B$, $x\notin A$. Ergo $x\in B\setminus A$.

Since cases $2.1$ and $2.2$ are exhaustive, $x\in B\setminus A$. Therefore $x\in (B\setminus A)\cup(A\setminus B$) and so $x\in A\Delta B$.

Since cases $1$ and $2$ are exhaustive, $x\in A\Delta B$. Therefore if $x\in A$ iff $x\notin B$ then $x\in A\Delta B$. Since $x$ is arbitrary, $\forall x\Bigr((x\in A\ \text{iff}\ x\notin B)\rightarrow x\in A\Delta B\Bigr).$

Ergo $\forall x\Bigr(x\in A\Delta B\ \text{iff}\ (x\in A\ \text{iff}\ x\notin B)\Bigr)$. $Q.E.D.$

Is my proof valid$?$ I would also appreciate a simpler proof.

Thanks for your attention.

Answer 1

Your proof is correct but here are some remarks.

In case 1 of $(\rightarrow)(\rightarrow)$, saying "suppose $x\in A$ and $x\in A\setminus B$" is redundant since if $x\in A\setminus B$ then it must be in $A$. The same comment can be made about case 2. Phrasing them this way also obscures that these cases are exhaustive of the assumption $x\in (A\setminus B)\cup (B\setminus A)$.

The same comments can be made about the separate cases in $(\rightarrow)(\leftarrow)$. Put together, this suggests that you shouldn't separate $(\rightarrow)$ this way. Rather:

Start from your assumption $x\in (A\setminus B)\cup (B\setminus A)$, which suggests two natural cases: either $x\in A\setminus B$ or $x\in B\setminus A$. Now, within these two cases, try to prove all at once that "$x\in A$ iff $x\not\in B$" holds. Rather than being too formal with unpacking "iff" as two separate implications, it might be more illuminating to think of an "iff" as stating an equivalence of truth values. In other words, if I have two statements $P$ and $Q$, and I can prove both are true, then "$P$ iff $Q$" follows immediately since"$P$ iff $Q$" is equivalent to "$P$ and $Q$ are both true, or both false". So for, example, if you assume $x\in A\setminus B$ then by definition this means $x\in A$ and $x\not\in B$. So "$x\in A$ iff $x\not\in B$" is true. A similar thing happens if I assume $x\in B\setminus A$, but in this case I get that "$x\in A$ iff $x\not\in B$" is true because both "$x\in A$" and "$x\not\in B$" are false.

Like I said in my comments, this is all covering the same ground, but looking at things this way makes the proofs shorter and perhaps more natural for someone else to read. You have posted many similar questions from this textbook, and I think your work has shown that you have a good understanding of unpacking logical implications and Boolean connectives. So I think it would be natural for you to now think more deeply about the proof structure and try to find more natural or readable approaches, as compared to the cut and dry algorithmic approach of breaking everything the finest parts with subcases and so forth.

For the $(\leftarrow)$ direction, I will just repeat what I said in the comments. Assume that for all $x$, $x\in A$ iff $x\not\in B$. Now take an arbitrary $x$. The goal is to show $x\in A\triangle B$. Given the definition of $\triangle$, it's perfectly reasonable to first eliminate the trivial case $x\in A\setminus B$ as you have done. But, going back to the view of "iff" that I discussed before, we can think of our assumption on $x$ as saying that the truth value of "$x\in A$" is the same as the truth value of "$x\not\in B$". So this suggest that more natural cases would be to look at the truth value of "$x\in A$". If $x\in A$ then $x\not\in B$ (by our iff assumption), so $x\in A\setminus B$ by definition. On other hand if $x\not\in A$ then $x\in B$ (by our iff assumption), so $x\in B\setminus A$ by definition.