First order Logic - Open and closed Formula

Question

I have this open formula in First Order Logic:

∃x p(x,y) or q(x)

The fact is i dont understand how is it different from

∃x ( p(x,y) or q(x))

For me, they are the same. In fact, in the first formula, once i find the x that satisfies p(x,y), i have to use it also for q. Infact p and q share the same variable x. Once it's defined for p , it's also defined for q. Why am i wrong ?

It depends on the semantcis. How are defiend the "truth conditions" for $\exists x P(x,y) \lor Q(x)$ ? In some case, an open formula $\varphi(x)$ is defined to be true in an interpretation only when every its instance is true. — Mauro ALLEGRANZA, Feb 06 '18 at 12:00
In the first formula, the "two $x$" are not necessarily the same; the formula is equivalent to: $\exists z P(z,y) \lor Q(x)$. — Mauro ALLEGRANZA, Feb 06 '18 at 12:08
Maybe, i havent clearly understand what a open formula is. I searched in various sites, but i wasnt able to find anything useful. Can you help me ? — Tantaros, Feb 06 '18 at 12:09
Under the usual interpretations, the first is equivalent to $\exists z(P(z,y))\vee Q(x)$, while the second is not. — Malice Vidrine, Feb 06 '18 at 12:10
Woah ! This is insane. Why is it standard to use two 'x' variables even if they can represent at the same time different values ? — Tantaros, Feb 06 '18 at 12:28
Because one is free, and thus represent in a given context, a "certain value (or object)" while the other is bound, and $\exists x$ does not "represent" anything: it asserts that "there is something...". — Mauro ALLEGRANZA, Feb 06 '18 at 12:36
See the related post: definition-of-the-bound-and-free-variable as well as the post: when-do-free-variables-occur ? what-is-the-intuition-behind-them ?. — Mauro ALLEGRANZA, Feb 06 '18 at 12:38
In your first formula, $x$ is used as both a bound and a free variable. I don't know if that is allowed in your system, but to avoid confusion, you should use another free variable, something like: $\exists x:[p(x,y)]$ or $q(z)$. I like to use the letters a-j for bound variables and k-z for free variables. Then your first statement would be $\exists a:[p(a,y)]$ or $q(z)$. And your second statement would be $\exists a:[p(a,y)$ or $q(a)]$. — Dan Christensen, Feb 06 '18 at 15:40

Mauro ALLEGRANZA · Answer 1 · 2018-02-06T13:07:32.533

For simplicity, get rid of $y$ and consider:

$\exists x (Px \lor Qx)$.

Consider an interpretation with domain the set $\mathbb N$ of natural numbers, and interpret the predicate $Px$ with the property $x<0$ and the predicate $Qx$ with the property $x=0$.

In this case, the formula amounts to:

$\exists x ((x < 0) \lor (x=0))$

that is true in $\mathbb N$.

Now consider the formula:

$\exists x (x < 0) \lor (x=0)$,

and consider a variable assignment $v: \text{Var} \to \mathbb N$ such that $v(x)=1$.

In this case, the formula means:

"either there is a (natural) number which is less than $0$ or $1=0$"

that is false in $\mathbb N$.

What does a statement with a free variable, like e.g. $x=1$, mean?

We need a "context" (technically called: variable assignment function), i.e. a way to assign a "temporary meaning" to the free variables.

We can compare a free variable to a pronoun of natural language.

To assert "$x$ is red" is the same as "it is red": its meaning depends on what the context assigns to "it".

Consider the example where I'm at my desk: on top if it there are two pens: one red and one blue, and no books.

If I'm asserting:

"there is an object on my desk such that (either it is book or it is red)",

my assertion is true: on my desk there is something red.

Consider now the different sentence:

"either (there is an object on my desk that is a book) or it is red".

What does it mean ? It depends on what "it" denotes. If I'm pointing with my finger at the blue pen, the sentence is false.

First order Logic - Open and closed Formula

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