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Let $X$ be any set, and let $P(x)$ and $Q(x)$ be propositions which are defined for any $x\in X$. Say that we know that $\forall x \in X P(x)$ and that $\forall x \in X P(x) \implies Q(x)$. I want to know if it is valid to from this conclude $\forall x \in X Q(x)$, and if so which logical rules would allow this.

The reason why I am unsure whether this is valid, is because if $X$ is an infinite set, we can't form a proof of $\forall x \in X Q(x)$ by for each $x \in X$ separately that $Q(x)$ is true, and concatenate these proofs, because that would be an infinite proof and I thought that all proofs are finite. Also, this reply to one of my posts claimed one could prove something for every explicit natural $n$, and if it false for all $n$. However, what makes me suspect that the above reasoning is valid, is that we argue in a way shared across all $x \in X$ when we prove $Q(x)$ for all $x \in X$, whereas in cases where we were able to prove a statement for every explicit element of a set but couldn't prove it for all elements in a set, perhaps we argued in a different way for each element of the set when we had an explicit proof for that element.

So is that reasoning valid, and if so what formal law of logic justifies it?

Princess Mia
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    Yes, you can prove this using the rules for $\forall$ and $\implies$. Please try to do it yourself in your favourite deduction system for first-order logic: it is an instructive exercise. – Naïm Camille Favier Aug 05 '24 at 07:58

1 Answers1

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We can't form a proof of $\forall x \in X Q(x)$ by for each $x \in X$ separately that $Q(x)$ is true, and concatenate these proofs, because that would be an infinite proof and I thought that all proofs are finite.

Short response: Yes, very true. But we can form a proof of $Q(a)$ where $a$ is an arbitrary element of $X$. And that's enough to establish $\forall x \in X Q(x)$. That's the essence of the crucial "universal quantifier introduction" (aka "universal generalisation") rule, pivotal to first order logic.

Let me cut'n'paste'n'edit an earlier answer of mine which might help to fix ideas:

Consider the informal argument 'Everyone loves pizza. Anyone who loves pizza loves ice-cream. So everyone loves ice-cream.' Why is that valid?

Roughly: Pick someone, whoever you like. Then, s/he loves pizza. And so s/he loves ice-cream. But sh/e was arbitrarily chosen. So everyone loves ice-cream.

This informal argument can be spelt out with plodding laboriousness with explicit commentary like this:

  1. Everyone loves pizza. (That's given)

  2. Anyone who loves pizza loves ice-cream. (That's given too)

  3. Take some arbitrary person, call her Alice. Then Alice loves pizza (From 1)

  4. If Alice loves pizza, she loves ice-cream (From 2)

  5. Alice loves ice-cream (From 3, 4, by modus ponens)

  6. But Alice was an arbitrary representative person, so what applies to her applies to anyone: so everyone loves ice-cream.

Now consider the formal analogue of this proof (using $F$ for 'loves pizza" etc.)

  1. $\forall xFx\quad\quad\quad\quad$ Premiss

  2. $\forall x(Fx \to Gx)\quad$ Premiss

  3. $Fa\quad\quad\quad\quad\quad$ From 1., Univ. Instantiation

  4. $(Fa \to Ga)\quad\quad$ From 2., Univ. Instantiation

  5. $Ga\quad\quad\quad\quad\quad$ From 3, 4 by MP

  6. $\forall xGx\quad\quad\quad\quad$ From 5, Univ. Generalization

(UG on $a$ is legimitate as $a$ appears in no assumption on which (5) depends, so $a$ can be thought of as indicating an arbitrary representative member of the domain.)

So note, that we need here some symbol playing the role of $a$, not a true constant with a fixed interpretation, not a bound variable, but (as common jargon has it) a parameter which stands in, in some sense, for an arbitrary element of the domaain.

Now, in some syntaxes, variables $x$, $y$ etc. only ever appear bound, as part of quantified sentences. And parameters are typographically quite distinct, $a$ and $b$, etc. This is the usage in Gentzen, for example.

But another tradition (more common, but not for that reason to be preferred), we are typographically economical, and re-cycle the same letters as both true variables and as parameters. In other words, we allow the same letters to appear both as "bound" variables and as "free" variables. Then the formal proof will look like this:

  1. $\forall xFx\quad\quad\quad\quad$ Premiss

  2. $\forall x(Fx \to Gx)\quad$ Premiss

  3. $Fx\quad\quad\quad\quad\quad$ From 1., Univ. Instantiation

  4. $(Fx \to Gx)\quad\quad$ From 2., Univ. Instantiation

  5. $Gx\quad\quad\quad\quad\quad$ From 3, 4 by MP

  6. $\forall xGx\quad\quad\quad\quad$ From 5, Univ. Generalization

This is a superficial difference, however: the role of the free (unbound) variable is as a parameter. And we've seen, even in informal arguments, we use expressions like "s/he" or even "Alice" as parameters. Looked at in that light, there should be no mystery about why we need parameters in a formal logic too. And in one syntax, unbound variables play the role of parameters.

Obviously, the same basic points apply if you want to be using restricted quantifiers ($\forall x \in X$ vs $\forall x$)

Peter Smith
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