2

Starting at 7:52 of this video, the professor presents a proof of the following theorem: for any infinite set of well-formed formulas $\Sigma$ such that $\Sigma \vDash \alpha$, where $\alpha$ is a w.f.f., there exists a finite subset $\Sigma_0 \subset \Sigma$ such that $\Sigma_0 \vDash \alpha$.

This is how I understood his proof (with comments of mine in bold):

Since $\alpha$ is a w.f.f., it involves only finitely many propositional variables, so there are only finitely many truth assignments in its truth table.

Isn't this already an issue? There are only finitely many truth assignments whose domains are the set of just those propositional variables that appear in $\alpha$. But in the larger context of an arbitrary infinite $\Sigma$, whose formulas could collectively involve infinitely many propositional variables, there could be infinitely many truth assignments for each combination fo truth values of those variables that appear in $\alpha$.

Let $V_i$ be a truth assignment for which $V_i(\alpha) = \mathrm{F}$. Therefore it cannot be the case that $\Sigma$ is satisfied under this assignment, since $\Sigma \vDash \alpha$, which implies there must be some $\alpha_{ij} \in \Sigma$ such that $V_i(\alpha_i) = \mathrm{F}$. For each $V_i$, pick one of these $\alpha_{ij}$ and rename it as $\alpha_i$. We claim that $\Sigma_0 = \{ \alpha_i: for all $i$ \}$ is a finite set such that $\Sigma_0 \vDash \alpha$. To see why, note that $\Sigma_0 \vDash \alpha$ if an only if $\Sigma_0 \cup \{ \neg \alpha \}$ is not satisfiable.

Assume by way of contradiction that $\Sigma_0 \cup \{ \neg \alpha \}$ is satisfiable. Then there exists a truth assignment $V$ such that $V(\neg \alpha) = \mathrm{F} \implies V(\alpha) = \mathrm{F}$. At this point, he asserts that any such $V$ is one of those $V_i$ referenced earlier (those $V_i$ for which $V_i(\alpha) = \mathrm{F}$ --- "we are in one of those rows of the truth table of $\alpha$ where $\alpha$ is false"). Since we constructed $\Sigma_0$ in such a way that for every $\alpha_i \in Sigma_0$ at least one $V_i$ is such that $V_i(\alpha_i) = \mathrm{F}$, so $\Sigma_0$ isn't satisfied by these $V_i$, which contradicts our assertion that $\Sigma_0 \cup \{ \alpha \}$ is satisfiable, proving the theorem.

What stands out as an issue to me: In the case that $\Sigma$ involves infinitely many different propositional variables, there would be infinitely many truth assignments. In particular, there would be infinitely many truth assignments for any combination of truth values of those variables that appear in $\alpha$. By looking at just one of them and then picking one $\alpha_i$ for $\Sigma_0$ in the way described, we can't be sure that the other truth assignments that assign the same truth values to the variables in $\alpha$ would assign $\mathrm{F}$ to $\alpha_i$, breaking the proof. If $\Sigma$, despite being infinite, involved only finitely many propositional variables, then it would be possible to look at every single truth assignment $V_i$ for which $V_i(\alpha) = \mathrm{F}$ --- since there would be only finitely many, but possibly more than just $2^n$, where $n$ is the number of propositional variables in $\alpha$ --- and include in $\Sigma_0$ an $\alpha_i \in \Sigma$ for which $V_i(\alpha_i) = \mathrm{F}$ for each of these $V_i$. This would ensure that we are in fact "in one of those rows of the truth table of $\alpha$ where $\alpha$ is false".

So it looks to me like what was really proved is a version of the theorem where $\Sigma$ involves finitely many different propositional variables. Is this really the case?

jvf
  • 491

2 Answers2

0

Here is a slightly easier way of thinking about this, in which we rely on the fact that propositional logic is deductively sound and complete.

Suppose $\Sigma$ is an infinite set of well-formed formulas of propositional logic such that $\Sigma \models \alpha$ for some propositional formula $\alpha$. By the completeness theorem for propositional logic, $\Sigma \vdash \alpha$. This is witnessed by a formal deduction $D$ in whatever formal system you've proved the completeness theorem for (say, a Hilbert system or a natural deduction system). Crucially, $D$ will be a finite object: a sequence or tree of formulas. Let $\Sigma_0$ be the set of all $\sigma \in \Sigma$ which appear as axioms in $D$. $\Sigma_0$ is a finite subset of $\Sigma$, and clearly $\Sigma_0 \vdash \alpha$. Finally, by the soundness theorem for propositional logic, $\Sigma_0 \models \alpha$.

Benedict Eastaugh
  • 2,883
0

I think you are correct, and the proof works only when $\Sigma$ involves finitely many variables (but then it's not very interesting, because there are only finitely many non-equivalent formulas in $\Sigma$). To see this, consider this version: let allow only assignments which assign T to finitely many variables. The compactness theorem fails in this case: for $\Sigma = \{x_0 \vee x_1, x_0 \vee x_2, \ldots\}$ and $\alpha = x_0$ we have that any allowed assignment that satisfies $\Sigma$ also satisfies $\alpha$ (because any assignment assigns F to some $x_n$, and as it has to satisfy $x_0 \vee x_n$, it has to assign T to $x_0$). But for any finite subset of $\Sigma$ there is an allowed assignment that satisfies this subset but not $\alpha$. However, I don't see any part in the proof that are affected by what is allowed set of assignments.

The strange part is that author talks about $V_i$ as an assignment for only variables of $\alpha$, but to apply it to formula from $\Sigma$, it has to be an assignment for all variables. And then it's possible to find assignment $V'$ s.t. $V'(\alpha) = F$, but $V'$ differs from all $V_i$ (on variables not in $\alpha$).

To prove this result, we need to somehow significantly use what assignments are allowed. See, for example, https://math.stackexchange.com/a/3393894/659499. In this prove, it's used that we can make set of all assignments into a compact topological space, s.t. for any formula, assignments that make it true, form a clopen set. In my example above, the set of allowed assignments isn't compact (under usual topology on assignments; and if we take different topology, s.t. it will become compact, then set of assignment satisfying formula will sometime fail to be clopen).

mihaild
  • 15,368