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In this video, we're going to talk about

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a proof, but instead of

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typing the proof…well discovering it

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we're going to read a proof, 
that I had just written.

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So the statement that we're 
going to prove is the following:

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so we're going to prove that A intersect 
B…so let me make this a little bigger.

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We're going to prove that
A intersect B is equal to

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A minus - or A set difference 
with A set difference B. Okay?

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So again we have a proof involving 
set equalities, so in order to show this,

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we're going to first show

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that the left-hand set is contained 
inside the right-hand set, and then

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we'll show that the right-hand set 
is contained inside the left-hand set.

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So first, we show that A 
intersect B is contained inside

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A set difference with A set difference B.

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So to do this, we start with an
element from the left-hand set.

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Let x be inside A intersect B,

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and what does that mean? 
So now we unwind the definition,

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so if [x] is inside the intersection, 
then it must be inside both the sets,

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right? So then x is in A and x is in B.

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By definition, what do we know? 
Well if x is in A and x is in B,

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then we know that x can't be in 
the set difference of A minus B,

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right, because it's in the intersection…

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or the set difference. So

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what does that tell us? Well however, 

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we already know that x is inside A

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and we know that x is not 
inside A set difference B,

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and so by definition again we have that

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x must be inside A set
difference with A minus B.

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Again that’s…this is just definition pushing.

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So you just have to remember 
what these definitions mean,

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right? A intersect B means that 
each element of A is contained in B,

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and A set difference with B means that

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x is an element of A and x is not

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an element of B. That's what 
the set difference set is.

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Now we try the reverse inclusion, right?

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So let x be inside A set difference 
with bracket A set difference B.

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By definition, we know that this means that, 
okay well we know that this means that

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x is in A and that x is not inside

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A set difference B.

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That is the definition of 
x belonging to this set.

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Now the next question is for x to not be 
inside this set, what does that mean? Well

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this means one of two things,

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it either means that x is not inside A,

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in which case would be done, 
but we already know that's false

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so we know that this part can't be true,

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but there's an “or”, right? So either 
x was not inside A or x is inside A,

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and x is inside B, right, because 
if it's inside both of these sets,

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then it's not in the set difference
because it's common, okay?

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Since we know the first part is false, 
right, we already have that x was in A,

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what can we conclude? Well we conclude 
that the second part has to be true, so then

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thus x is an element of A 
and x is an element of B,

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and then hence, by definition, 
x must be inside the intersection.

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That shows the reverse containment,

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so now we've shown
the double containment, 

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and so maybe I should make 
a concluding sentence…

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therefore…

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the claimed statement 
is true. So let’s just…

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easy to do on a computer 
when you can copy and paste.

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So what should our concluding sentence 
be? Therefore A intersect B is equal to

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A set difference with A set difference B.

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Okay, hopefully this helped.
Thank you very much.