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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).

Squirtle
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  • How do you get $p - q = -(p - q)$? – Rob Arthan Jan 22 '15 at 22:07
  • Every Boolean ring is essentially a Boolean algebra and visa-versa; this is true by construction/definition. A Boolean ring is a ring with unit in which every element is idempotent ($p^2=p$). – Squirtle Jan 22 '15 at 22:11
  • maybe $p\subset q$ is a better way to think of $p \Rightarrow q$ - $p$'s "truth zone" is within $q$'s "truth zone". – Joffan Jan 22 '15 at 22:13
  • I don't know if I agree with this statement, because $p\implies q$ is defined as $p'\vee q$ which if we are working in the scenario of sets traslates as $(p^c\cup q)^{\perp\perp}$ Where $^\perp$ means the complement of the closure. If $p$ and $q$ are regular open sets then this is simply $p^c\cup q$ and I definitely don't see the connection then. – Squirtle Jan 22 '15 at 22:21
  • We're asserting that $(p^c\cup q)=U$, which is the same as $p \subset q$ – Joffan Jan 22 '15 at 22:38
  • If you are quoting Halmos accurately, then his $p - q$ cannot mean subtraction in the Boolean ring corresponding to the Boolean algebra (subtraction and addition are the same in a Boolean ring, so subtraction is commutative, but implication is not commutative in a Boolean algebra). How does Halmos define "-"? – Rob Arthan Jan 22 '15 at 22:46
  • $p-q$ is defined as $p\wedge q'$, and $p+q$ is defined as $(p-q)\vee(q-p)$. With ordinary sets, the difference is set subtraction and addition is the symmetric difference. – Squirtle Jan 22 '15 at 23:08
  • So $p - q \not= -(p - q)$ in general and your concern about "two duals" is not a problem. – Rob Arthan Jan 23 '15 at 00:34
  • $p-q$ DOES equal $-(p-q)$ in general, I quoted an exact proof from page two of the text. – Squirtle Jan 23 '15 at 00:41

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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 $(p\Longrightarrow q) = \{1,3,4\}$.

Michael Hardy
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  • I didn't downvote this answer. It was helpful although not at all what I was expecting, so I am waiting to see if I'll get any other responses. Kinda surprised the question got down voted too, some people .... – Squirtle Jan 23 '15 at 20:49
  • I like the example, thanks. +1 – Squirtle Jan 24 '15 at 05:53