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PROFESSOR: So now we come
to an interesting variant

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of ordinary induction
called strong induction,

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and here's how it works.

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With strong induction, just
as with ordinary induction,

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you prove the base case P of 0.

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You're trying to prove for all
nP of n, so you prove P of 0,

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but now in order to prove P of
n plus 1 in the inductive step,

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assuming P of n with
ordinary induction,

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with strong induction you
can assume not just P of n

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but you can assume
P of 0, P of 1.

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All the properties-- that
all the numbers up through n

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have the property.

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And from this, of
course, you could

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conclude that that
everything has the property

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that for all mP of m.

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Now, an intuitive way
to justify this is you

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think about it the way
that induction works is

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you start off at 0, and
then you make a step to 1,

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and you make another step to 2,
and you make another step to 3,

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and the induction step
going from n to n plus 1

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justifies each of those steps.

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By the time you get
to n and you have

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to prove you can
get to n plus 1,

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you've already been
through 0 1 up to n.

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You might as well take
advantage of that fact.

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Instead of only remembering
that you're at the n step,

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you might as well remember
that you got there.

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That's an intuitive
hand-wavy argument

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which can be justified
formally in a way that will

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emerge in the next segment.

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So let's hold off on that
and just bite the bullet

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and accept this as a
basic principle of math

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that we're going to
live with and use.

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As an application of it,
let's prove something

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that we've already proved by
well ordering and, in fact,

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strong induction and well
ordering are closely related,

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as we'll also discuss later.

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So let's prove that using
$0.03 and $0.05 stamps that you

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can get any amount of
postage greater than

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or equal to $0.08 stamps,
and I'm going to prove this

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by strong induction,
with the induction

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hypothesis P of n that says I
can form n plus $0.8 stamps.

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Clearly, if I can
prove for all nP of n,

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then I've proved that I can get
for every amount greater than

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or equal to $0.08 stamps.

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And let's do the base case.

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Well, the base case, P of 0.

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Can I make $0.8 stamps?

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Sure, $0.03 and $0.05.

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That's that one, and that's OK.

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For the inductive
step, I have to get

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m-- I'm allowed
to assume, rather,

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that I can get m plus
$0.8 for any m from n

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down to 0, instead
of just assuming

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that I can get n plus $0.8
to get n plus 1 plus $0.08.

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I can assume any amount less
than what I'm aiming for,

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so I may as well assume that I
can get any amount of postage

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from $0.08 up to n plus
$0.08, and my objective then

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is to get n plus 1 plus
$0.08, namely n plus $0.09.

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So I have to prove that for all
n greater than or equal to 0,

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I can get n plus $0.09,
assuming I can get from $0.08

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to n plus $0.08.

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Well, that's not too hard to do.

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The inductive step
is actually going

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to break up into
a couple of cases,

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depending on the value of n.

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I have to prove n plus $0.09 for
all n, so suppose n equals 0,

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I have to get $0.09.

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Well, three $0.03.

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If n is 1, I have to get 1
plus $0.09 or $0.10, two $0.05.

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So those cases are disposed of.

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So now my job is to get
n plus $0.09, where n

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is greater than or equal to 2.

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Well, the nice
thing about n being

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greater than or equal to 2 is
that if I subtract 2 from it,

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it's a smaller number, and
it's still not negative,

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and that means that I can get
that amount plus $0.08 stamps.

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So I'm in this nice situation
where I, by strong induction,

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I can get n minus 2
plus $0.08 stamps.

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There they are.

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And how do I get to n plus 9?

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Well, it's easy.

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I add a $0.03
stamp, and you have

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n plus $0.09, which completes
the proof of the induction

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case, and the whole
theorem is proved.

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We can conclude then
that it works for all n

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and that you can indeed
get n plus $0.08 using

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$0.03 and $0.05 stamps
for all of them.

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So much for that example.

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All right, let's look
at another example.

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This is a game that we
used to play in class.

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You start off with a stack
of blocks, say 10 blocks,

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and you're allowed
to make a move that

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consists of splitting the
stack into two smaller stacks.

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So if the stack has
height a plus b,

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you can split it into a
stack of height a and a stack

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of height b, and you get
a score for that move.

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The score is a times b.

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And then you keep doing that
until you can't make anymore

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moves.

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That is, when all you have
left are stacks of height one,

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which you can't split anymore,
and then your overall score

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is the total that you
got for all the moves

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that you made until that point.

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Now, when we played
this in class,

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we would have
students competing,

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and they would try
various strategies.

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So one strategy-- the simplest
strategy, maybe not the best,

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but the simplest
strategy, would be

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to start off with a stack
of 10, so you take one of,

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and that leaves you with
a stack of one and nine.

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Your score is nine.

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Then you take another one
off of the stack of nine,

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and you're left with
a one and an eight.

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Your score is eight, and so
on, and you can see, in fact,

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if took one-at-a-time process
then your score with a stack

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of height n would be n
minus 1 plus n minus 2,

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down to 2 plus 1.

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Another strategy that
might be sort of more

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in the spirit of
computer science

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would be to keep
splitting in two.

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So for example, if you
had a stack of height 10,

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you could split
it into two fives,

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and then you take
one of the fives

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and split it into
three and a two,

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and then you'd split
the two into two ones,

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and so on, splitting as
evenly as you can each time,

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and it seems like it might
be a better strategy.

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And, again, we
would have students

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try various strategies,
and guess what?

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They all came in in
a tie, and that's

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what we're going to prove now.

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Every way of unstacking n
blocks gives the same score.

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Well, what score?

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Well, we know that the score for
the simple strategy of taking

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one block off at a time is
the sum from 1 to n minus 1,

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and that has a nice formula--
n times n minus 1 over 2,

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so we can formulate our claim
that no matter how you play

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the unstacking game with
the stack of size n,

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your final score will be
n times n minus 1 over 2,

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and we're going to prove
this by strong induction,

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with this statement--
call it claim of n--

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is going to be the
induction hypothesis.

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That's what we're
trying to prove.

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Well, let's start
in the usual way.

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The base case is n equals 0.

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Well, you might be bothered.

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That's no blocks.

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Well, let's see what happens.

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With no blocks, the score is 0
because there's nothing to do,

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and indeed the formula that
is alleged to be your score

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comes out to be 0, so the
base case n equals 0 works.

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Let's continue.

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For the inductive case, I have
to assume that the given score

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formula holds for all
stacks of height n or less,

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and I have to
prove that it holds

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for a stack of height n plus 1.

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That is, that an n
plus 1 stack score

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is n plus 1 times n over 2.

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Well, how shall I do that?

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Well, I'm going to split the
inductive case into two cases.

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It turns out that I need
to prove that c of n plus 1

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holds, assuming c of n for n and
less than n but, in particular,

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let's just deal with
the case that n plus 1

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is 1, the smallest value
it could have and knock

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that one off separately.

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Namely, if the stack
is of height one,

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again my score is 0, because
there's no move to make,

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and the formula
still evaluates to 0.

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So in the case that n plus 1 is
1, I've proved the claim at n

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plus 1, which I was
obligated to do for the base

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case-- for the inductive step.

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Well, the other case in the
inductive step is that n plus 1

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is greater than 1.

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This is the interesting
one, because now it's

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possible to make a move.

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So since n plus 1
is greater than 1,

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it's two or more blocks.

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I can make a move
into two stacks

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that are both of positive.

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So suppose I do that?

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Suppose I split the
stack of size n plus 1

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into an a stack and a b stack,
where a and b sum to n plus 1.

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And what's my score
going to be then?

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Well, my score on
that move that I make,

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where I split into the a
stack and the b stack is a b,

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and the rest of
the game consists

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of playing as well as
I can on the a stack

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and as well as I
can on the b stack,

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but a and b are
smaller than n plus 1.

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They're less than
or equal to n, which

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means that by strong induction,
I know that no matter how

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I play on the a stack,
I'm going to wind up

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with this score a
times a minus 1 over 2.

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No matter how I
play on the b stack,

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I'm going to wind up with
b times b minus 1 over 2,

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so that means that my
score on a plus b stack

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is going to be this formula, ab
plus a times a minus 1 over 2

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plus b times b minus 1 over 2.

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So you simplify
that to organize it

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so it's a plus b times a plus
b minus 1, which is exactly

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n plus 1 times n
over 2, which is

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what we were trying to prove.

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We've proved C event plus 1.

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The inductive step is
complete, and indeed we've

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proved that no matter
how big the stack is,

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your score comes out the same.