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I'm trying to prove that it's not the case that $\Sigma \vdash \bigwedge_x F_x$, where $\Sigma= \{FA_1, FA_2, FA_3, \ldots , FA_n,\ldots\}$

I started to prove by contradiction. So assume that $\bigwedge_x F_x$ is derivable. Then by the soundness theorem $\Sigma \models \bigwedge_x F_x$.

I'm thinking that the contradiction has to do with the finiteness condition for $\models$, and since we're considering all $x$ in the one-place predicate $F$, that we cannot get a finite set out of $\Sigma$. So it's not consistent, which is the contradiction.

Is this the correct way to prove this?

Michael Hardy
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Jon Harris
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2 Answers2

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You could observe that $$\{Fa_1, Fa_2, Fa_3, \ldots Fa_n \ldots \} \nvDash \forall xFx$$ by remarking that there are obviously interpretations which makes all the premisses true and the conclusion false. [Take the domain to be people, $F$ to have as extension the set of great philosophers, and all the $a_j$ to denote Aristotle ...] And hence, by the soundness of the relevant first-order deductive relation, $$\{Fa_1, Fa_2, Fa_3, \ldots Fa_n \ldots \} \nvdash \forall xFx.$$

Peter Smith
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OK, the domain is all non-negative integers and your assertions are about all positive integers, asserting that each has a reciprocal within the field of rational numbers. But it's not true that every non-negative integer has a reciprocal within the field of rational numbers, so that's a counterexample.

Or the domain is the set of all positive integers, and each statement in your sequence of assertions says something about an even integer: it says that it's even. That's a counterexample.

Michael Hardy
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