# Show that $\Sigma\vDash p_1\vee p_2\vee…\vee p_n$

This is Exercise 3.27 from Derek Goldrei's, Propositional and Predicate Calculus : A Model of Argument.

Suppose that, for each $$i\in\mathbb{N}$$, $$p_i$$ is a propositional variable. Let $$\Sigma$$ be a set of sentences of the propositional calculus. Suppose that all truth assignments which satisfy $$\Sigma$$ make at least one $$p_i$$ true. Show that for some $$n\in\mathbb{N}$$, $$\Sigma\vDash p_1\vee p_2\vee...\vee p_n.$$

I am aware that an answer exists here. I want to instead attempt to prove the contrapositive:

Suppose that $$\Sigma\vDash p_1\vee p_2\vee...\vee p_n$$ for no $$n\in\mathbb{N}$$. This is the same as saying $$\Sigma\nvDash p_1\vee p_2\vee...\vee p_n$$ for all $$n\in\mathbb{N}$$. This trivially means that for some truth assignment, $$v_n$$, which satisfies $$\Sigma$$, that $$v_n(p_1\vee p_2\vee...\vee p_n)=F$$, for all $$n\in\mathbb{N}$$. (The $$v$$ does not necessarily have to be the same for all $$n$$'s and is hence subscripted by $$n$$.) By the truth table for $$\vee$$, we conclude that $$v_n(p_1)=v_n(p_2)=...=v_n(p_n)=F$$ for all $$n\in\mathbb{N}$$. But this says that one truth assignment, $$v_{n\to\infty}$$ which satisfies $$\Sigma$$ makes none of $$p_i$$, $$i\in\mathbb{N}$$ true (i.e. the one in the limit that $$n\to\infty$$), and we are done proof by contrapositive.

Is this a valid proof method in this situation?

• What you must do is to argue from the existence of all $\nu_n$ that there is a valuation $\nu$ such that $\nu(p_i)=F$ for all $i$ and $\nu$ satisfies $\Sigma$. – Andrés E. Caicedo Sep 30 '18 at 20:11
• @AndrésE.Caicedo Crudely, is this not the $v_n$ in the limit that $n\to\infty$? – quanticbolt Sep 30 '18 at 20:13
• I don't know what that means. – Andrés E. Caicedo Sep 30 '18 at 20:59
• I mean that $v_{\infty}$ is the truth assignment which satisfies $\Sigma$ but not any $p_i$. But I don't think I can use infinities like that in logic. – quanticbolt Sep 30 '18 at 21:01
• You talk of "the truth assignment", as if it is unique, which is completely unjustified as far as I can tell – Andrés E. Caicedo Sep 30 '18 at 21:03

No, that's not a valid argument. The problem is when you say

But this says that one truth assignment which satisfies $$\Sigma$$ make none of $$p_i$$, $$i\in\mathbb{N}$$ true

That's not what it says. You have for each $$n$$ a valuation $$\nu_n$$ which makes $$\Sigma$$ true but no $$p_i$$ true for $$i\le n$$, but you haven't cooked up a single valuation $$\nu$$ which makes $$\Sigma$$ true but no $$p_i$$ true for any $$i$$ at all.

For example, maybe $$\nu_1\models p_2$$, $$\nu_2\models p_3$$, $$\nu_3\models p_4$$, ...

Now we want to say that somehow we can "smoosh together" the $$\nu_i$$s to get a $$\nu$$ making $$\Sigma$$ true and making each $$p_i$$ false. But how?

(HINT: use the compactness theorem ...)

(We can sometimes think of this as "taking the limit" of the $$\nu_n$$s, but that takes a bit of work to make non-problematic. One approach to doing so is to introduce ultrafilters - and this is basically avoiding using the compactness theorem by employing its proof.)

• What if I define my $v$ as follows. $v(p_i)=v_i(p_i)$? Then clearly $v$ satisfies $\Sigma$ but also $v(p_i)=F$ for all $p_i$. – quanticbolt Sep 30 '18 at 20:34
• @quanticbolt "clearly $\nu$ satisfies $\Sigma$" Why? That's essentially what you're trying to prove. – Noah Schweber Sep 30 '18 at 20:45
• @quanticbolt What do you mean "$\nu$ is only ever $\nu_i$"? Suppose $\nu_i$ makes each $p_j$ false for $j\le i$ but each $p_k$ true for $k>i$. In what sense is $\nu$ only ever $\nu_i$? – Noah Schweber Sep 30 '18 at 20:48
• hmmm. I'm sensing that this is where the compactness theorem comes in, but I'm not exactly sure how to apply it. I know that it says $\Gamma$ is satisfiable iff every finite subset of $\Gamma$ is satisfiable. So in this case, is my $\Gamma=\Sigma$? – quanticbolt Sep 30 '18 at 20:50
• @quanticbolt Not quite. It's not $\Sigma$ alone that you want to satisfy. What's the whole theory you want your $\nu$ to satisfy? It's $\Sigma$, but also something about the $p_i$s ... – Noah Schweber Sep 30 '18 at 20:51