Re: logical implication with quantifiers

Well...I'm still a bit confused. Take a concrete example to make sure
I get it right:
E - there exists
A - for every
C(x,y), D(x) are binary and uniary predicates respectively

For a sentence Ey(C(x,y))->D(x)
equivalent to (Ay~C(x,y)) v D(x)

Now because of this:
((Ay)~Cy) v Bx

(Ay)(~Cy v Bx) because y is not free in Bx

and since y is free in D(x), I can say (Ay)(~C(x,y) v D(x)) and hence
it gives me (Ay)(C(x,y)->D(x))?


Ken Pledger wrote:
In article <1144806794.005267.259050@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"MoeBlee" <jazzmobe@xxxxxxxxxxx> wrote:

Kevin Chill wrote:
Sorry, I meant to say Ex(Ax)->Bx (not Ex(Ax->Bx)). Now if I want a
disjunction form, I should flip the quantifier, right?

You're using 'A' as a predicate letter, not as a quantifier, right? So
I'll change your 'A' to 'C', since I'm using 'A' as a quantifier. So I
take it you're asking about:

ExCx -> Bx ....


This notation is perfectly correct, but it uses "x" as both a bound
and a free variable in the same formula. Logicians regularly do that,
but I think mathematicians more often use an extra letter to make things
intuitively easier. The above formula is equivalent to

((Ey)Cy) -> Bx or ((Et)Ct) -> Bx or etc.

In the first of those notations I'll take some of MoeBlee's
transformations a bit further. Here are some equivalents to your
formula.

((Ey)Cy) -> Bx

(~(Ey)Cy) v Bx

((Ay)~Cy) v Bx

(Ay)(~Cy v Bx) because y is not free in Bx

(Ay)(Cy -> Bx).

That last version may be what you were trying to find.

Ken Pledger.

.

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