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Boolean Algebra Again?
There has always been a hint in the air that what we have been doing with propositions mirrors what we did with Boolean algebras. Let's see what evidence we have so far on this connection.
 Propositional logic is based on 0 and 1
 There are operations $p + q$, $pq$ and $p'$
 $p + p' = 1; (p')' = p; (pq)'= p'+ q'; $and $(p + q)'= p'q'$;
All of these appear in Boolean algebra. To be sure that we have a Boolean algebra all we have to do is to make sure that the axioms of Boolean algebra hold for propositions (page 16). Let's look at a few for a start.

Axiom 1
$B$ contains at least two elements. Propositional logic contains at least true and false. These hover in the background just as on and off do for switching circuits. 
Axiom 2
$B$ is closed under + and $\times$. That is for all $x$ and $y$ in $B, x + y$ and $x \times y$ are in $B$. For propositions $p$ and $q$, $p + q$ and $ pq$ are both propositions. 
Axiom 3
+ and $ \times$ are commutative. That is for all $x$ and $y$ in $B, x + y = y + x$ and $x \times y = y \times x$. So we have to show that $p + q = q + p$ and $pq = qp$ are tautologies. We do the first in Table 9.
$p$  $q$  $p+q$  $p+q$ 
0  0  0  0 
0  1  1  1 
1  0  1  1 
1  1  1  1 
From here we can see that $(p + q) \to (q + p) $ and $(q + p) \to(p + q)$. So + (either or) in propositional logic are the same.
Questions
 Prove that $pq = qp$.
 Prove that the remaining axioms are also true.