I'm reading Halmos's Lectures on Boolean Algebras. The title is a definition and he then also defines $p\iff q= (p\implies q)\wedge (q\implies p)$. Then the following:
The source of these operations suggests an unintelligent error that it is important to avoid. The result of the operation $\implies$ on the elements $p$ and $q$ of the Boolean algebra $A$ is another element of $A$; it is not an assertion about or a relation between the given elements $p$ and $q$. (The same is true of $\iff$.) It is for this reason that logicians sometimes warn against reading "$p \implies q$" as "$p$ implies $q$" and suggest instead the reading "if $p$, then $q$".
How do I make sense of this, I believe this statement to be true; however, we do in practice interpret "$p \implies q$" as "$p$ implies $q$" (in fact I had to type \implies). Or put another way, when we do say "$p$ implies $q$" (or for that matter, "If $p$, then $q$") about any mathematical statement, what is the simplest connection to Boolean algebra? Another issue I have is this notion of duality:
The operations $\implies$ and $\iff$ would arise in any systematic study of Boolean algebra even without any motivation from logic. The reason is duality: the dual of $p-q$ is $q\implies p$, and the dual of $p+q$ is $p\iff q$.
I see that $p+q=q+p$ so the dual of this is also $q\iff p$, I would normally consider this as the same as $p\iff q$ (however, perhaps there's more really going on?). But $p-q=-(p-q)=q-p$ which has dual $p\implies q$ (following the quote above about the first dual); so $p-q$ has two duals (if I understand this correctly)!
EDIT:
Two of its most surprising consequences are that (1) a Boolean ring $A$ has characteristic $2$ (that is, $p+p=0$ for every $p$ in $A$), and (2) a Boolean ring is commutative. For the proof, compute $(p+q)^2$, and use idempotence to conclude that $pq+qp=0$. This result implies the two assertions, one after another, as follows. Put $p=q$ and use idempotence to get (1); since (1) implies every element is equal to its own negative, the fact that $pq=-pq$ yields (2).
I can't draw a Venn diagram here. Imagine two overlapping circles inside a rectangle that is the "universe" --- the typical simple Venn diagram. Let the first circle be $p$ and the second $q$. Color in all of $q$ and all of the area outside of both circles, so that the only part that is not colored in is the part of $p$ that is outside of $q$. Then the shaded area is $p\Longrightarrow q$. It contains all points that are not counterexamples to "if $p$ then $q$".
Suppose the "universe" is $\{1,2,3,4\}$ and $p=\{1,2\}$ and $q=\{1,3\}$. Then $(p\Longrightarrow q) = \{1,3,4\}$.