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PROFESSOR: So now
we're going to talk

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about the concepts of
validity and satisfiability,

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which take on some extra
interest in the context

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of predicate calculus.

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So let's remember for
propositional validity,

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if you have a propositional
formula with variables taking

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the truth values ranging
over true and false,

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then a formula is
valid when it's

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true for all possible
truth values.

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So here's an example that
P implies Q or Q implies P.

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And you can check that for the
four possible environments of P

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and Q-- true/false
values of P and Q,

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this OR will come
out to be true.

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Well, for predicate formulas
there's a bunch of things

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that I need to
give value to that

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are more complicated
than just truth values.

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In particular, I will say that
a predicate calculus formula

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is valid when it's true for all
possible domains of discourse

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that the variables range over.

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There's a technicality that it
has to be non-empty, but aside

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from that all possible domains.

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And whenever you have
a predicate mentioned

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in the formula, in order to
know whether the formula is true

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or not, I need to know
what that predicate means.

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So a formula is valid if it
comes out true no matter what

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the predicate means.

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Let's look at a concrete
example to get a grip on this.

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Here is a valid formula
of predicate calculus.

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It's mentioning
predicates P and Q.

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It's of the form
of a proposition

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because it's saying something
about every possible z

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in the domain and every
possible x and every possible y.

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The only thing that we need
to know to make sense out

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of this formula, to figure
out whether or not it's true,

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is what's the domain that
x, y, and z range over

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and what exactly
do P and Q mean?

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Well, I want to
argue informally.

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Let's just look at what
this is saying together.

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What this is saying is
suppose that for everything

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in the domain, both
property P of z and P of Q.

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In other words, everything
in the domain have property P

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and property Q. Well,
that certainly implies

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that everything in the
domain has property

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P because they have
both properties.

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And also, everything
in the domain

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has property Q because
everything has both properties.

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So when you say it
that way, the sense

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that this is a fundamental
logical fact that

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doesn't depend on what P and
Q mean or what the domain is.

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It's just a fact about the
nature of the meaning of the

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for all universal quantifier and
the connectives [? AND then ?]

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implies.

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That's how we figure
out that this is valid.

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Well, let me go one
level in more detail

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to say again what I
just said informally

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and try to be a little
bit more precise and clear

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about a reason why
this formula is valid.

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So suppose I wanted to prove
that the formula is valid.

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Well, it's an implication.

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So the proof strategy--
there it is written again--

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the proof strategy is
I'm going to assume

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that the if part, the
left-hand side of the implies

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or hypothesis, is true.

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That is, that for every z, P
of z holds and Q of z holds.

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And then I'm going to try
to prove, based on that,

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that the consequent
holds, namely

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that the right-hand side for all
x P of x and for all y Q of y

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holds.

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OK.

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How am I going to do that?

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Well, so here's the
formula written just

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to fit on the line with the
concise mathematical symbols.

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The upside down V means AND,
and the arrow means implies.

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And we want to try to prove that
this is valid a little bit more

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carefully.

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Well, the strategy,
as I said, is

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to assume that the
left-hand side holds.

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Well, what's the
left-hand side say?

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It says that for every z, Q
of z holds and P of z holds.

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That means that for every
possible environment that

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assigns a value to
z, Q of z and P of z

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both come out to be true.

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Well, suppose that the
environment assigns the value c

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to z, where c is some
element of the domain.

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Then what this means is that in
that environment, Q of z and P

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of z is true, which means
that Q of c and P of c holds.

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But Q of c holds and P
of c holds, so Q of c

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certainly holds all by itself.

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All right.

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So now we're in an
interesting situation

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because we just proved
that Q of c holds.

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And we know nothing and
have assumed nothing

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about c except that it's
an element of the domain.

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c could have been any
element of the domain,

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and we've managed to
prove that Q of c holds.

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So it follows that, in
fact, we have really proved

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that for every x Q of x holds.

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Now, that step of saying I
proved it for Q of a given

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element without making any
assumptions about the given

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element except that it's in
the domain and therefore I can

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conclude that it holds
for all domain elements,

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[? it's  a ?] very natural and
plausible and understandable

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rule.

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And it's a basic axiom of
logic called UG-- Universal

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Generalization.

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We'll come back
to it in a minute.

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Anyway, I've just proved
that for all x Q of x holds.

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And by a completely symmetric
argument, for all y P of y

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holds.

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And having proved both for all
x Q of x and for all y P of y,

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clearly the AND holds.

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And I've just proved
that the right-hand side

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of this implication is true
given that the left-hand side

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is true.

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Now, having called
this proving validity,

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let me immediately
clarify that this is not

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fair to call a proof because
the rules of the game

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are really murky here.

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This theorem, you
could read it as saying

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that universal quantification
distributes over AND

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is one of these
basic valid formulas

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that is so fundamental
and intelligible that it's

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hard to see what
more basic things

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you are allowed to assume
when you're proving it.

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And this proof really
isn't anything more

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than translating upside down A
and the AND symbol into English

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and using ordinary intuitive
rules about for all and AND

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and using that in the proof.

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So this is a good way to
think about the formula

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to get an understanding of it.

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But it's not right
to say that it's

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a proof because we haven't
been exactly clear about what

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the proof rules are.

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And with this kind of really
fundamental valid fact,

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it becomes a quite
technical problem

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to decide what a
proof is going to be.

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What's fair to assume and
what's fair not to assume?

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It would actually be
perfectly plausible to take

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this as an axiom and
then prove other things

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as a consequent of it.

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Anyway, going on, let's look at
this just for cultural reasons.

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We're never going to actually
ask you to do anything with it.

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But the universal
generalization rule UG

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would be stated this way as
a deduction rule in logic.

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The stuff over the bar
means if you've proved this,

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then you can conclude you've
proved the stuff below the bar.

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So what this is
saying is if you've

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proved P of c for
a constant c, then

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you can deduce that for
every x P of x holds.

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And this is providing
that c does not

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occur in any other
part of the predicate P

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except where you're talking
explicitly about it.

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It's hard to be more
precise about that for now.

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Don't worry about it.

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But the idea is you're not
supposed to assume anything

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about c other than
it's in the domain

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and that it has property P,
and you can then conclude

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that everything has property P.

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So let's look at
a similar example

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where it is possible
to prove something.

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Namely, I can prove that
something's not valid.

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So here's a
similar-looking formula.

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This one says that for every
z if P of z holds OR Q of z

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holds, then for every x P of
x holds OR for every y Q of y

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holds.

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And this one we're going
to show is not valid.

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Let's think about
it for a minute.

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What it's saying is if
everything has either property

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P or property Q, that implies
that everything has property P

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or everything has property Q.
Well, when you say it that way,

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it's clearly not the case.

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But let's go one level more
precise and lay that out.

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What I'm going to
do is convince you

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that it's not valid by giving
you a counter-model where

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I choose an interpretation.

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I choose a domain of
discourse and predicates

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that P and Q are going
to mean over that domain

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and that make the left-hand
side of this implication true.

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And then I'm going to show you
that the right-hand side is not

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true.

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And that means
that in that domain

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with those interpretations
of P and Q,

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this implication fails
so it's not valid.

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So I need to make the
left-hand side true

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and the right-hand side false.

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Well, I'm going to choose
the domain of discourse

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to be the simplest one
that will make this false,

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namely let's let the domain of
discourse just be the numbers

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1 and 2.

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And let Q of z be the predicate
that says z is 1 and P of z

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be the predicate
that says z is 2.

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Well, is the
left-hand side true?

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Yeah, because the only things
there are in the domain

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are 1 and 2, and so clearly
everything in the domain

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is either 1 or 2.

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So the antecedent is true.

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On the other hand, is
everything in the domain,

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does it satisfy P?

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Is everything in the
domain equal to 2?

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No, 1's not equal to 2.

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What about, is everything
in the domain equal to 1?

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Is it true that for
all y Q of y holds?

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No, 2 is in the domain,
and it's not equal to 1.

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And so we have
found exactly what

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we wanted, a
counter-model which makes

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the left-hand side
of the implies true

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and the right-hand side
of the implies false.

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Let me close with just one
more example of a valid formula

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that we can talk through.

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This is the version
of De Morgan's law

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that works for quantifiers.

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Remember De Morgan's
law was the thing that

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said that the negation
of P or Q was the same

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as not P and not
Q. And remembering

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that the connection between
universal quantification,

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[? an ?] AND, and existential
quantification, [? an ?] OR,

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it turns out that by the
same kind of reasoning,

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De Morgan's law
comes out this way.

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It says that if it's
not true that everything

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has property P,
that's possible if

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and only if there's something
that doesn't have property P.

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And so that's what
De Morgan's law is.

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It's another thing you
could take as an axiom,

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or you could try one of these
hand-waving proofs about.

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But I think I've said
enough to give you

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that example of another
interesting valid formula,

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and we'll stop with that.