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PROFESSOR: Let's
define a few familiar

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and standard operations on sets.

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So here's a picture
of two sets A and B,

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where the idea is that
the circle represents

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the points in A. The
other circle represents

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the points in B. The overlapping
area, this lens-shaped region,

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are the points that
are in both A and B.

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And the background
are the points that

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are in neither A and B.

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So this sort of
a general picture

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allows you to classify
points with respect to And B,

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and it's called a Venn diagram,
in this case for two sets.

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It's still useful
for three sets.

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It gets more complicated
for four sets.

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And after that point, they're
not really very useful.

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But a lot of the
basic operations

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can be illustrated
nicely in terms

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of the Venn diagram
for two sets,

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and that's what
we're about to do.

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So the first operation is union.

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It's the set of points
shown here in magenta.

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It's the set of points that are
in either A or B, all of them.

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And so if we were
defining this in terms

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of set-theoretic notation
or predicate notation,

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the union symbol-- the
U is the union symbol.

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So A union B is defined to be
those points x that are in A OR

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are in B.

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And you can already begin to
see an intimate relationship

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between the union operation
and the propositional OR

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connective.

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But don't confuse them.

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If you apply an OR to
sets, your compiler

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is going to give
you a type error.

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And if you apply union to
propositional variables,

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your compiler is also going
to give you a type error.

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So let's keep the
propositional operators

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and the set-theoretic
operators separate

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[? and ?] clearly
distinct even though they

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resemble each other.

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All right.

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Next basic operation is
intersection where, again, it's

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the points that are
both in A and B,

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the points in common, which
are now highlighted in blue.

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So the definition of
A intersection B--

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we use an upside-down union
symbol for intersection--

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it's the set of points
that are in A and are in B.

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So let's stop for
a minute and make

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use of the similarity between
the set-theoretic operations

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and the propositional operators.

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Let's look at a set-theoretic
identity, which I claim

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holds no matter what sets A,
B, and C you're talking about.

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And we're going to
prove it by making

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the connection between
set-theoretic operations

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and propositional operators.

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And so let's read the thing.

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It says that if you take A
union the set B intersection

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C, that's equal to the
set A union B intersected

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with A union C. Now, let's
not think through yet

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how to make this an
intuitive argument.

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It's going to really crank
out in an automatic way

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in a moment.

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But we can remember
it as saying that you

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can think of this as union
distributing over intersection.

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So if you think of
union as times here

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and intersection
as plus, then we've

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got a rule that says
that A times B plus C is

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A times B plus A times C.

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Now, it's also true that if
you reverse the role of union

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and intersection, you get
another distributive law

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that AND distributes over
union, but never mind that.

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Let's just look at this one.

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We're trying to prove the
distributive law for union

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over intersection.

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How shall we prove it
just from the definitions?

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Well, the way we're
going to do it

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is by showing that the two
sets on the left-hand side

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and the right-hand side have
the same set of elements.

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Namely, if I have
an element x that

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appears in the set described
on the left-hand side,

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then that point is in
the right-hand side.

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And it's an if and only if.

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So that says that the left-hand
side and the right-hand side

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expressions defines sets
with the same set of points.

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This holds for all x.

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And it turns out
that the proof is

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going to follow by analogy
to a propositional formula

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that we're going to make
use of in the proof.

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That was a propositional
equivalence

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that we proved in an
earlier talk, namely

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that OR distributes over AND.

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So P OR Q AND R is equivalent
to P OR Q AND P OR R.

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So you can see this
equivalence in purple

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has the same structure as the
set-theoretic equality in blue,

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except that union's
replaced by OR,

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intersection's replaced
by AND, and set

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variables A, B, C is replaced
by propositional variables

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P, Q, R.

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So let's just remember
that we've already

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proved this propositional
equivalence,

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and we're going to make use of
it in the middle of this proof

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that these two sets are equal.

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So again, we said we
were going to prove

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the two sets are
equal by showing

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they have the same points.

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So here's the proof.

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It's going to be a lovely if and
only if argument the whole way.

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So looking at the left-hand
side, a point x is in A union

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B intersection C by
definition of union

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if and only if x is an A OR
x is in B intersection C.

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I've just applied the
definition of union there.

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OK.

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Now, let's look at
this expression.

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x is in B intersection
C. That's the same

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as x is in B AND x is
in C, again, just using

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the definition of intersection.

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And now I have a propositional
formula involving OR and AND

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and the basic assertions about
sets of x is a member of one

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of those A's, B's, C's.

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Now, at this point,
I can immediately

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apply my propositional
equivalence

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and say that the assertion x is
an A OR x is in B AND x is in C

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holds if and only if
this expression, x is

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an A OR x is in B AND x is in
A OR x is in C. Why is that?

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Well, I'm just invoking the
propositional equivalence.

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Let's look at it.

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That if I think of the x
is in A as proposition P--

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and let's replace all the
x [? over ?] A's by P--

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and I think of x is in B as
a Q and x is in C as an R,

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then I can see that the first
set-theoretic assertion has

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the form of P OR Q AND R.

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And I can transform it by
the propositional equivalence

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into P OR Q AND P OR
R. And then remember

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what P and R are to get back to
the set-theoretic membership,

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basic membership assertions.

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So now we've just
proved that x is

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in A OR x is in B AND
x is in A OR x is in C.

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And that's if and only if it
was in the left-hand side set.

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Well, now I'm going to
go back the other way.

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Namely, this OR, that
x is in A OR x is in B,

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is the same as saying
that x is in A union B,

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likewise here just by applying
the definition of union.

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And this assertion that
x is in this set AND x

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is in this set is the
same as saying that x

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is in their intersection.

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And I've completed
my proof, namely

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the point that was in the
left-hand side if and only

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if it's in the right-hand side.

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You have to remember that
that was the right-hand side

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of the identity.

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So this is a general
method actually,

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where you can take any
set-theoretic equality

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involving union and
intersection and the operations

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of difference and complement
that we'll talk about

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in a moment, and we can convert
any such set-theoretic equality

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into a propositional equality
or a propositional equivalence

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so we can check that the
propositional assertion is

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an equivalence.

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And from that, using this method
of converting the membership

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statements in the set
expression into a propositional

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combination, we can check,
and automatically check,

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any kind of set-theoretic
identity involving

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union, intersection, and minus.

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And that, in fact, is the
way that automatic engines

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like Mathematica can prove
these set-theoretic identities.

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So let's just for the record
put down that last operation.

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The difference
operation is the set

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of elements that are
in A AND not in B.

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So we'd write it as A minus
B is the set of points that

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are in A AND not in B.
and it's Illustrated

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by this region that's
highlighted in orange.

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And a special case of the minus
operation or the difference

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operation is complement.

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When you know the overall domain
that you expect all your sets

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to be part of, then you
can define a complement

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to be everything that's not
in A-- the set of x such

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that x is not in A, where x
is understood to be ranging

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over some domain of discourse.

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So if we're going to
picture that, we're

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looking at the whole orange
region, all of the stuff that's

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not in A if we think
of the whole slide

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as representing the
domain of discourse D.