Satisfaction of a Quantified WFF

In case my explanation in lecture of the definition of satisfaction for a quantified wff wasn't clear, here it is at greater length (for a finite case, which may make it easier to understand, and can be done "wolog", i.e. "without loss of generality").

Suppose our domain, D, contains the following items:

d1, ..., d200

Suppose our language has these variables:

x1, ..., x100

The definition of satisfaction for a universally quantified wff is:

, v. iff , for all that differ from at most on v.

E.g., suppose v is the variable: x50
and is the wff: "x50^2 = x50*x50"

Consider the set of all variables: {x1, ..., x50, ..., x100}

Consider what set of members of D we get by applying a variable assignment to all the members of this set:

{(x1), ..., (x50), ..., (x100)}

Let's say that this set = {d1, ..., d50, ..., d100}.

Now consider all that differ from at most on x50. Suppose they are 1,...,300.

(It may be that 301,...,1000 differ from on other variables, too, so we don't consider these at all!)

Note that our above is one of these 300 (because all the differ at most on x50, meaning that they might not differ on x50 at all). For convenience, let's say that our above is 1.

So consider what all these do to the set of all variables; they map them into sets of elements of D as follows:

1: {1(x1), ..., 1(x50), ..., 1(x100)}
.
.
.
300: {300(x1), ..., 300(x50), ..., 300(x100)}

Note, in particular, that 1 gives us: {d1, ..., d50, ..., d100}. What about the others? Well, each differs from (or 1) at most on what it does to x50, so we get the following:

1: {d1, ..., d49, d50, d51, ..., d100}
2: {d1, ..., d49, d1, d51, ..., d100}
.
.
.
300: {d1, ..., d49, d77, d51, ..., d100}

I.e., they are all alike on the first 49 variables and on the last 50 variables, but they can vary wildly on good old x50.

Now, according to the definition, if all of these satisfy "x50^2=x50*x50", then our original will satisfy x50[x50^2=x50*x50].

But to decide if a given satisfies "x50^2=x50*x50", we can ignore what it does to everything except x50.

(We can do this for 2 reasons: First, they don't differ on the other variables. Second, our wff only contains that one variable, so that's the only one we care about.)

On x50, each differs. Does each one satisfy "x50^2=x50*x50"? Sure! Here's the proof:

d50^2=d50*d50
d1^2=d1*d1
...
d77^2=d77*d77

So: all the that differ from at most on x50 satisfy x50^2=x50*x50, so satisfies x50[x50^2=x50*x50].

And similarly, "mutatis mutandis" (as mathematicians say; i.e., changing what needs to be changed), for the existential quantifier.