Proving open formula

Question

Here $x,y$ are just real numbers. If given an open formula like $\forall x (y=5x)$, how do we prove that the resultant closed formulae are false for all possible choices of $y$? The way I think to do it is to write $\forall x \forall y (y=5x)$, then set $x$ to be $1$ giving $\forall y (y=5)$ which is false.

Answer 1

Since this question was inspired by this one and this one about the truth-value(s) of open formulas, I'm going to give a somewhat long-winded answer.

Deductive systems. There are a variety of deductive systems for the predicate calculus. Without getting into the weeds, they all share the same goal. Namely, if all models $M$ satisfying a bunch of formulas $\Gamma$ (notation: $M\models\Gamma$) also satisfy $\varphi$, then the deductive system should provide a purely syntactic way to deduce $\varphi$ from $\Gamma$ (notation: $\Gamma\vdash\varphi$).

A "subgoal": one should be able to convert natural-language mathematical proofs into formal proofs.

Inference rules. This article lists four rules for the classical predicate calculus. Let's look at two in more detail: universal generalization (UG), and universal instantiation (UI). You'll find them both at work in the example proof in this article.

Universal instantiation is easier to understand. A natural-language proof might say, "We know (or have assumed) that $\varphi(x)$ holds for all $x$, so in particular it must hold for this element $t$." Formally $t$ would be a term and $x$ a variable, so (UI) takes the form "Infer $\varphi(t)$ from $\forall x\varphi(x)$". (The article lists some restrictions, needed to avoid pitfalls.)

Universal generalization shows up in natural language proofs in this way. One starts by laying out the hypotheses $\Gamma$. Next: "Let $x$ be some element." Then some reasoning, concluding that $\varphi(x)$ holds. Finally: "But since $x$ was an arbitrary element, $\forall x \varphi(x)$ is true (still assuming $\Gamma$)."

The article mentions the essential restriction that $x$ not occur free in $\Gamma$, since then $x$ would not be an arbitrary element. (E.g., if our assumptions included the formula $\exists n(x=2n)$, then $x$ would have to be even.)

Harking back to your earlier question, you can now see the usefulness of the convention mentioned in my answer.

Proof that $\forall x(y=5x)$ is false for all choices of $y$ (in $\mathbb{R}$). The natural-language argument goes like this: "If this holds for all $x$, it must hold in particular for $x=1$. So we've shown that $y=5$: that's the only choice of $y$ making the claim true. But likewise, if it holds for all $x$, it must hold for $x=-1$, and so $y=-5$. We've now shown that $y=5$ and that $y=-5$, which is a contradiction."

A formal proof would be pretty long, but you can see the skeleton. $\forall x(y=5x)$ is our hypothesis, added on to $\Gamma$. ($\Gamma$ consists of the field theory axioms plus some stuff specific to $\mathbb{R}$.) Using UI twice, plus $\Gamma$, you can show formally $y=5$ and also $y=-5$, or finally $5=-5$. This is a contradiction, so the rules of propositional calculus allow you to conclude $\neg\forall x(y=5x)$. We have now "discharged" the hypothesis. That is, the conclusion has been shown to follow just from $\Gamma$. So using UG, we conclude $\forall y(\neg\forall x(y=5x))$.

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Now, what about the semantic interpretation of these inference rules?

The goal of a deduction system, as I said, is to capture semantic implications. Suppose $M\models\Gamma$ implies $M\models\varphi$; we want to have $\Gamma\vdash\varphi$. Initially assume $\Gamma$ and $\varphi$ have no free variables.

The deductive-semantic correspondence is clearest, I think, with so-called natural deduction systems. Here each line of a deduction is written $$\varphi_1,\ldots,\varphi_n\vdash\psi$$ with the semantics "$M\models\varphi_1,\ldots,\varphi_n$ implies $M\models\psi$". Again, this is with no free variables.

What if one of the $\varphi$'s or $\psi$ contains a free variable? In that case, the simplest approach is to expand the model $M$ by adding a new constant to the language, say $c_y$ for a free variable $y$. An expanded model $(M,c_y)$ assigns a value to $c_y$. There will be many expansions $(M,c_y)$ of a given model $M$. (Same idea for many free variables.)

To say $M\models\psi(y)$ means that all of the expansions $(M,c_y)\models\psi(c_y)$. Note that $\psi(c_y)$ is a closed formula. And $\varphi_1(y),\ldots,\varphi_n(y)\vdash\psi(y)$ means that for all models $(M,c_y)$, if $(M,c_y)\models\varphi_1(c_y),\ldots,\varphi_n(c_y)$ then $(M,c_y)\models\psi(c_y)$.

Finally, the UI rule takes the form "From $\varphi_1,\ldots,\varphi_n\vdash\forall y\varphi(y)$ infer $\varphi_1,\ldots,\varphi_n\vdash\varphi(t)$", and the UG rule the form "from $\varphi_1,\ldots,\varphi_n\vdash\varphi(y)$ infer $\varphi_1,\ldots,\varphi_n\vdash\forall y\varphi(y)$", subject to the proviso that $y$ doesn't occur free anywhere in $\varphi_1,\ldots,\varphi_n$.