Drake, Seven Axioms of the Algebra of Events On page 3 of Drake Fundamentals of Applied Probability he lists The Seven Axioms of the Algebra of Events.
$1. A \cup B = B \cup A \\
2. A \cup (B \cup C) = (A \cup B) \cup C \\
3. A\cap( B \cup C) = A \cap B \cup A \cap C \\
4. (A')' = A \\
5. (A\cap B)' = A' \cup B'\\
6. A\cap A' = \phi \\
7.A \cap U = A  $
I have two questions.
First while Drake states that his choice is not unique is there a reason that he selected these seven and is seven the minimum possible number of axioms with which the Algebra of events can be expressed.
Second he further states that any other relation in the algebra of events can be proved directly from these seven axioms with no additional information and in fact in the chapter problem section asks the student to do so. Does the author mean that all the other set relations in the Algebra of Events can be proved by direct manipulation of the seven axioms with out resorting to the method of $x \in A$
For example, proofs of $A \cup A' = U$ are given in this posting
Proof of union of a set and its complement is equivalent to a universe
but these proofs use the method of $x \in A$ and not direct manipulation of Axiomatic statements as required by Drake. How would I prove $A \cup A' = U$ or $A \cup A = A$ using only direct manipulation of Drake's Seven Axioms of the Algebra of Events?
 A: remark
From 1-7 we deduce that $U$ is unique.  Suppose
$$A \cap U = A  \text{ for all }A\tag7$$
$$A \cap V = A  \text{ for all }A\tag{7'}$$
We claim $U = V$.
Proof. From $(7)$ we get $V \cap U = V$.
From $(7')$ we get $U \cap V = U$.
Then apply $(1)$ to get $V \cap U = U \cap V$, and conclude $U = V$.
A: Let us answer first the second part of your question and the first part of your question.
Let us prove $A \cup A' = U$ and $A \cup A = A$, using only direct manipulation of Drake's Seven Axioms of the Algebra of Events (as it is asked in the question).
Consider the seven axioms:
$1. A \cup B = B \cup A \\
2. A \cup (B \cup C) = (A \cup B) \cup C \\
3. A\cap( B \cup C) = (A \cap B) \cup (A \cap C) \\
4. (A')' = A \\
5. (A\cap B)' = A' \cup B'\\
6. A\cap A' = \phi \\
7.A \cap U = A  $
Let us go result by result .
a. From axioms $1$, $5$  and $4$, we have $A\cap B = B\cap A $.
$ A' \cup B' = B' \cup A' \textrm{ by } 1 \\
(A\cap B)' = (B\cap A)'  \textrm{ by } 5 \\ 
A\cap B = B\cap A     \textrm{ by } 4 $
b. We also have $U'=\phi$ and $\phi'=U$.
$  U\cap U' = \phi   \textrm{ by } 6 \\
U'\cap U = \phi   \textrm{ by item a. above}  \\
U'= \phi    \textrm{ by } 7  \\
U= \phi'    \textrm { by } 4 $
c. From axioms $6$, $5$, $4$ and $1$ you can now easily get  $A\cup A' = U$:
$A\cap A' = \phi  \textrm{ start in  } 6 \\
A'\cup (A')'= \phi' \textrm{ by } 5 \\
A'\cup A  = \phi'  \textrm{ by } 4 \\
A \cup A'  = \phi'  \textrm{ by } 1 \\
A \cup A'  = U  \textrm{ by item b. above} $
d. We have that $A \cup \phi = A$.
$ A' \cap U = A' \textrm{ by } 7 \\
(A'\cap U)' = A   \textrm{ by } 4 \\
(A')' \cup U' = A  \textrm{ by } 5 \\
A \cup U' = A   \textrm{ by } 4 \\
A \cup \phi = A  \textrm{ by item b. above} $
e. Now let us prove that $A \cap A =A$.
$ A \cap U = A \textrm{ by } 7 \\
A \cap (A\cup A') = A \textrm {by item c. above} \\
(A \cap A) \cup (A \cap A') = A \textrm{ by } 3 \\
(A \cap A) \cup \phi = A \textrm{ by } 6 \\
A \cap A = A \textrm{ by item d. above }  $
f. Let us prove now that $A\cup A  = A$
$ A' \cap A' = A' \textrm{ by item e. above }  \\
( A' \cap A' ) ' = A \textrm{ by } 4 \\
( A')' \cup ( A' ) ' = A \textrm{ by } 5 \\
A \cup A = A  \textrm{ by } 4 $
Now regarding the first part of your question.
The choice is not unique, of course. Consider, for instance,
$1. A \cup B = B \cup A \\
2. A \cup (B \cup C) = (A \cup B) \cup C \\
3. A\cap( B \cup C) = (A \cap B) \cup (A \cap C) \\
4. (A')' = A \\
5. (A\cap B)' = A' \cup B'\\
6'. A \cup A' = U  \\
7'. A\cup \phi = A $
From axioms $1$ to $7$ we have already proved $6'$ (see item c. above) and $7'$ (see item d. above).
Now using axioms $1$ to $5$, $6'$ and $7'$, the proof of item a. above remains unchanged because it uses only axioms $1$, $4$ and $5$. Then
b'.  Let us prove $\phi'=U$  and $U'=\phi$.
$  \phi\cup \phi' = U   \textrm{ by } 6' \\
\phi'\cup \phi = U   \textrm{ by item a. above}  \\
\phi'= U    \textrm{ by } 7'  \\ 
\phi= U'    \textrm { by } 4 $
Using the item b'. above and axiom $4$, it is easy to recover axiom $6$ from axiom $6'$ and axiom $7$ from axiom $7'$. So we can conclude that the set of axioms $1$ to $7$  is equivalent to the set of axioms $1$ to $5$, $6'$ and $7'$.
Which set to choose is more a matter of personal choice. Both sets are fine.
Remark: Regarding your question if the number of axioms is the minimum. Well, we can always reduce the number of axioms simply using "and", so we could replace, for instance, axiom $4$ and axiom $5$ by an axiom $4+5$ saying:
$ 4+5. \:\: (A')' = A \textrm{ and } (A\cap B)' = A' \cup B' $
Actually, any finite set of axioms is equivalent to a single (long) axiom.
On the other hand, disregarding the simple concatenation by "and", I would say that seven is probably the minimum number of axioms. The argument here is not a "rigorous proof", it is more an argument of intuition. Let us see it.
We have three operation ($\cup$, $\cap$, $'$)  and two constants ($\phi$, $U$). We should expect to have at least one axiom to define each of the operations and each of the constants. So we should have at least five axioms. But we also need axioms to define the interaction of the operations and that would bring the total of axioms to six or seven axioms.
Note that in Drake's axioms, he uses two axioms to describe union (axioms $1$ and $2$), but just one axiom to define intersection and how complement interacts with unions and intersections (axiom $5$).
A: What Drake calls the Algebra of Events is usually called Boolean Algebra. There are many ways to axiomatize Boolean Algebra, and Drake has given one of them.

First while Drake states that his choice is not unique, is there a reason that he selected these seven...

I don't see any reason to prefer this axiomatization to others. It's a little inconvenient since, as you noted, it's tricky to see why some basic equations follow from these axioms.

...and is seven the minimum possible number of axioms with which the Algebra of events can be expressed?

Wikipedia has a page on Minimal axioms for Boolean Algebra, but the axiomatizations discussed there work in a context with fewer basic operations (e.g., viewing $A\cap B$ as an abbreviation for $(A'\cup B')'$, or just using the Sheffer stroke). I don't know what the minimum number of equations is that axiomatizes Boolean Algebra with all its usual operations.

Second he further states that any other relation in the algebra of events can be proved directly from these seven axioms with no additional information...

Here's the idea of how to prove such a thing: If some equation is not provable from the axioms of Boolean Algebra, then there is an abstract Boolean algebra in which the equation does not hold - for example, the free Boolean algebra on the variables appearing in the equation. Now the Stone representation theorem says that every abstract Boolean algebra is isomorphic to an algebra of subsets of some set $U$. So there would be a counterexample to the equation obtained by substituting subsets of $U$ for the variables in the equation. Thus, if some  equation holds whenever the variables are interpreted by events (subsets of a set $U$), then the equation can be proved from the axioms of Boolean Algebra.
Now if you want to be sure that Drake's list of axioms fully axiomatize the Algebra of Events, instead of checking that every equation that holds of events follows from Drake's axioms (there are infinitely many such equations!), you just need to check it for each axiom in some known axiomatization of Boolean Algebra (like this one).
