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The proof to

$\Gamma \models \phi [\nu /\kappa ] \: \Rightarrow \: \Gamma \models \left ( \forall \nu \right )\phi$

(Provided that no sentences in $\Gamma\:$ have occurrences of $\kappa$ and also that $\kappa$ does not occur in $\phi$).

Goes as follows:

Suppose that $\:\Gamma \models \phi [\nu /\kappa ]$. For RAA, assume that it is not the case that $\:\Gamma \models \left ( \forall \nu \right )\phi$. Then there is some interpretation $\:\mathbf{I}$ that satisfies $\:\Gamma$ but not $\:\left ( \forall \nu \right )\phi$. Since $\kappa$ does not occur in $\phi$, there is some variant $\textbf{I}_{\kappa}$ of $\:\mathbf{I}\:$ with respect to $\kappa$ such that $\phi [\nu /\kappa ]$ is false under $\textbf{I}_{\kappa}$. But also $\kappa$ does not occur in any sentence of $\:\Gamma$. Since the only difference between $\:\mathbf{I}$ and $\textbf{I}_{\kappa}$ is what they assign to $\kappa$, and since $\:\mathbf{I}\:$ satisfies $\:\Gamma$, therefore $\textbf{I}_{\kappa}$ satisfies $\:\Gamma$. Thus, $\textbf{I}_{\kappa}$ satisfies $\:\Gamma$ but not $\phi [\nu /\kappa ]$, which contradicts the hypothesis that $\Gamma \models \phi [\nu /\kappa ]$. $\mathbf{Q.E.D.}$

My question is,

Why does it not matter what $\kappa$ is chosen for $\:\Gamma$? Specifically, I am not understanding the significance of the sentence,

Since the only difference between $\:\mathbf{I}$ and $\textbf{I}_{\kappa}$ is what they assign to $\kappa$, and since $\:\mathbf{I}\:$ satisfies $\:\Gamma$, therefore $\textbf{I}_{\kappa}$ satisfies $\:\Gamma$.

Does $\:\mathbf{I}\:$ satisfying $\:\Gamma$ implicitly mean that any arbitrary constant that does not occur in $\Gamma$ satisfies all substitution instances inside $\Gamma$? Nothing suggests that $\Gamma$ is satisfied for all interpretations so why doesn't any $\kappa$ affect it?

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  • $\begingroup$ The only way I can see it working is if there are no variables inside of $\Gamma$ to create any substitutions using $\kappa$, then variants of $\mathbf{I}$ with respect to any constant not occurring in $\Gamma$ would not matter. Does saying "$\Gamma$ is a set of Predicate Calculus sentences" imply that there are no variables in $\Gamma$? $\endgroup$ – skyfire Jan 12 '17 at 21:34
  • $\begingroup$ Yes; sentence means "closed" formula, i.e. a formula without free occurrences of variables. $\endgroup$ – Mauro ALLEGRANZA Jan 13 '17 at 7:37
  • $\begingroup$ The gist of the restriction is quite simple; consider violating it: we have in $\Gamma$ the sentence : $k=1$ and assume as $\phi$ : $v>1$. Of course, we cannot conclude that: $(\forall v)(v >1)$ because in $\mathbb N$ we have also $0$. $\endgroup$ – Mauro ALLEGRANZA Jan 13 '17 at 7:40
  • $\begingroup$ Thank you for the help and clarification. I think the above and the answer really cleared everything up for me. $\endgroup$ – skyfire Jan 13 '17 at 17:09
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Let's fix a first order language $\mathcal{L}$, where $\Gamma$ is a set of sentences in the language $\mathcal{L}$. Suppose $I$ is a model of $\Gamma$ satisfying $\lnot \forall v \phi(v)$. That is, there is $a \in I$ such that $I \models \lnot \phi(a)$.

If we expand the language to $\mathcal{L} \cup \{ \kappa \}$, where we add a new constant symbol, $\Gamma$ is still a set of sentences in this language. In order to make $I$ to be a model in this expansion, we need an interpretation of the constant symbol $\kappa$. So let's call $I_\kappa$ the interpretation $I$ with $\kappa$ interpreted as the element $a$. $I_\kappa$ satisfies the same $\mathcal{L}$-sentences as $I$, since it has the same universe and the same interpretations for any symbols in $\mathcal{L}$. In particular, if $I \models \Gamma$ then $I_\kappa \models \Gamma$.

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  • $\begingroup$ I think I understand your answer, you are essentially saying what I added in my comment. Albeit the language is slightly different. Essentially for any considered interpretation of the sentences in $\Gamma$, by the assumption, they are already interpreted satisfactorily for the language $\mathfrak{L}$ and adding a new constant to the language $\mathfrak{L}$ does not change the interpretation of the sentences inside of $\Gamma$. $\endgroup$ – skyfire Jan 12 '17 at 21:50
  • $\begingroup$ Right, the point is that $\Gamma$ does not put any restrictions on what the interpretation of the new constant symbol $\kappa$ could be. So you are free to pick any element of your model to be the interpretation of $\kappa$. $\endgroup$ – Athar Abdul-Quader Jan 13 '17 at 2:18
  • $\begingroup$ I should probably amend my first comment on this answer to say "For the interpretation that satisfies $\Gamma$ and not $\left ( \forall \nu \right )\phi $, adding a new constant to the language $\mathfrak{L}$ does not change the interpretation of the sentences inside of $\Gamma$. $\endgroup$ – skyfire Jan 13 '17 at 17:19

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