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In first order logic,is $(\text{True}\lor P(x))$ considered $\text{True}$ (where the variable $x$ is free) ?

One we can approach is as follows, since there is a $\text{True}$ in a dis-junction with $P(x)$ where $P(x)$ is a predicate, then irrespective of the what value $P(x)$ takes the first order logic (FOL) expression could considered to be $\text{True}$.

But what also bugs me in the above is that, if $x$ is free, then $P(x)$ is not actually a proposition and hence it shall not have any truth value associated with it. As such assuming $P(x)$ to be $?$ in this case, [$?$ is something unknown], is it quite right to apply a logical connective to it and get a truth value, such as $\text{True}\lor ? \equiv \text{True}$.

I mean, we could just write $Q(x): \text{True} \lor P(x)$, as $x$ is free, we cannot assign a truth value to the predicate $Q(x)$ , it is not a proposition...

Please help me.

Mauro ALLEGRANZA
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Abhishek Ghosh
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    Ultimately the idea "proposition = statement with definite truth value" doesn't really play well with the taxonomy which mathematical logic finds more useful: we distinguish between formulas and sentences based on whether there is a free variable, and a well-formed expression with free variables is a formula rather than a sentence even if it is true in all structures under all variable assignments. I think the following helps: we're shifting away from the issue of whether we have enough information to determine whether an expression is true - and towards the idea of definable sets. – Noah Schweber Nov 22 '21 at 17:11
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    (And definable functions and relations, more generally. See e.g. my answers here and here.) – Noah Schweber Nov 22 '21 at 17:16

2 Answers2

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P(x) is not actually a proposition and hence it shall not have any truth value associated with it.

It's more accurate to say that predicates typically have varying truth values.

$Q(x): \text{True} \lor P(x)$

we cannot assign a truth value to the predicate Q(x) , it is not a proposition...

Do you agree that the predicate $$(x=y\land y=z)\to x=z$$ is true? If so, then the predicate $Q(x)$ is likewise true.

We don't need to know the actual formulation of $Q(x)$'s left disjunct "True" to assert that it is a validity; likewise, we don't need to know $Q(x)$'s full formulation (its right disjunct is unspecified) to assert that $Q(x)$ is a validity.

ryang
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The issue is: are we approaching the question formally or intuitively?

From a formal point of view, truth can be ascribed to a formula when we specify an interpretation, i.e. a domain $D$, a "meaning" for the predicates, i.e. for a unary predicate $P$ a subset of the domain, and a "temporary" meaning to free variable, for example with a variable assignment function $s : \text {Var} \to D$.

Thus, assume for example an interpetation $\mathcal I$ with domain the set $\mathbb N$ of naturals, as meaning for predicate $P$ the property "$x \text { is Even}$", which means that $P^{\mathcal I} = \{ n \mid n \in \mathbb N \text { and } n \text { is Even} \}$, and as variable assignment the function $s(x)=3$.

Now we have all we need to compute $\mathbb N, s \vDash (\text T \lor Px)$, i.e. the truth value of the formula for the given interpretation.

In plain text we have to check if the statement "either (True) or (3 is Even)" is true in the domain of naturals.

According to the truth table for "or" the answer will be: YES, and the obvious fact is that this will hold irrespective of the value that we choose for $x$.

If instead we approach it from an intuitive point of view, a free variable acts as a pronoun. Thus, "$Px$" can be read as "it is (a) $P$".

The truth value depends of course on the meaning of $P$ but also on the way we "understand" the pronoun: are we pointing with our finger to something on the table when we utter the statement?

If so, we have simply to check if that object is "a P".

Mauro ALLEGRANZA
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