Added a note

Advanced/Definable Sets/parts/0 logic.tex

@@ -66,7 +66,7 @@ $\lnot A$ is the opposite of $A$, which is why it looks like a \say{negative} si
 \vspace{2mm}
 
 $A \rightarrow B$ is a bit harder to understand. Read aloud, this is \say{$A$ implies $B$.} \par
-The only time $\rightarrow$ is false is when $T \rightarrow F$. Think about it: why does this make sense? \par
+The only time $\rightarrow$ is false is when $T \rightarrow F$. This may seem counterintuitive, but it makes sense. Think about it. \par
 
 \problem{}
 Evaluate the following.
@@ -78,6 +78,24 @@ Evaluate the following.
 
 \vfill
 \pagebreak
+\begin{instructornote}
+	After the class has done a few definable set problems, you can try to provide some intuition for $\rightarrow$ with the following example.
+
+	\vspace{2mm}
+
+	Say we have the sentence $\forall x ~ (a \rightarrow b)$. \par
+	For example, take $\varphi = \forall x ~ ([x \geq 0] \rightarrow [\exists y ~ y^2 = x])$. \par
+	$\varphi$ holds whenever any positive $x$ has a square root.
+
+	\vspace{2mm}
+
+	If $(\text{F} \rightarrow *)$ returned false, statements like the above would be hard to write. \par
+	If $x$ is negative, $\varphi$ doesn't care whether or not it has a root. In this case, $\text{F} \rightarrow *$ must be true to avoid making whole $\forall$ false.
+
+	\vspace{2mm}
+
+	You can think of $[x \geq 0] \rightarrow b$ as a \say{sanity check} in a program: if $x$ isn't the kind of object we care about, return true and check the next one. If $x$ \textit{is} the kind of object we care about and $b$ is false, we have a counterexample to $[x \geq 0] \rightarrow b$, and thus $T \rightarrow F$ must be false.
+\end{instructornote}
 
 
 \problem{}