basic theoretical question regarding difference between a formula and a proposition

Question

i am trying to understand the difference between a formula and a proposition, so i have a basic theoretical question regarding the difference between a formula and a proposition:

let's look at if $\not \vDash \omega$ then $\vDash \lnot \omega$:

what would be the difference if $\omega$ was a formula or a proposition?

trying to understand the difference between $\omega$ being a formula or a proposition. if you can, please explain the difference so i can learn and understand.

thank you very much for your help

EDIT: the question is about propositional logic, not how we interpret formulas. i am trying to understand if in one case the statement might not hold true and in the other it will(if $\omega$ is a formula or if it is a proposition).

i am wondering if in any case, $⊭ω$ then $⊨¬ω$ might be wrong(not correct) for one(either proposition or formula), but if we say that $\omega$ is the other, then it would be correct. trying to understand the difference

EDIT2: the connection between a formula and a proposition according to my textbook: a formula will be called a proposition iff it doesn't have any free variable. that is, if all of the occurrences of the variables in the formula are bound variables.

Answer 1

Regarding propositional calculus, for every $p$ atomic we may have a truth valuation $v$ such that $v(p)=$ TRUE and a different truth valuation $v′$ such that $v′(p)=$ FALSE.

Thus, neither $\vDash p$ nor $\vDash ¬p$, because the example shows that both are not tautologies : a formula is a tautology if it is evaluated to TRUE by every truth assignment.

In propositional calculus we have tautologies : $p \lor \lnot p$, contradicitions : $p \land \lnot p$, and satisfiable formulas : $p \land q$.

It is not true, in general, that the negation of a formula that is not a tautology must be a tautology.

See $p \land q$ above; it is not a tautology : consider the truth assignment $v$ such that $v(p)=$ FALSE.

But also its negation : $\lnot (p \land q)$ is not Consider the truth assignment $v$ such that $v(p)=v(q)=$ TRUE.

In symbols :

$\nvDash p \land q \text { and } \nvDash \lnot (p \land q)$.

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For predicate logic, we have to take into account free variables, and the situation is similar.

A formula with free occurrences of variables, like e.g $(x=0)$, is satisfiable. Consider the domain $\mathbb N$ of natural numbers.

But also its negation : $\lnot (x=0)$ is satisfiable in the same domain.

For predicate logic, we have that $\vDash \varphi$ means that the formula $\varphi$ is valid, i.e. TRUE in every interpretation.

But if we consider the relation between the formula and a specific interpretation, we have that :

if $\varphi$ is a sentence, i.e. a formula with no occurrences of free variable, and $\mathcal M$ is an interpretation with domain $M$, then :

$\mathcal M \vDash \lnot \varphi \text { iff } \mathcal M \nvDash \varphi$.

Consider the sentence $\forall x (x=0)$ with the usual interpretation with domain $\mathbb N$.

We have that :

$\mathbb N \nvDash \forall x (x=0) \text { and } \mathbb N \vDash \lnot \forall x (x=0)$.

Note that the last one is equivalent to : $\exists x \lnot (x=0)$.