Do these two $\forall$-introduction rules agree with each other?

Asked Oct 04 '23 at 07:18

Active Oct 04 '23 at 07:18

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Enderton's An Mathematical Introduction to Logic says on p112:

$\alpha \to \forall x \alpha$ if $x$ does not occur free in $\alpha$.

Why is the requirement of $x$ not free in $\alpha$?

Ebbinghaus' Mathematical Logic says on p69:

5.5 Exercises (b4)

$ \frac{\Gamma \quad \vdash \quad \phi}{\Gamma \quad \vdash \quad \forall x \phi} $ if $x$ is not free in $\Gamma$.

Why does it require $x$ not free in $\Gamma$ not in $\phi$?

Do the two agree with each other?

asked Oct 04 '23 at 07:18

Tim

2

1st question: because $x=0 \to \forall x (x=0)$ is not true. – Mauro ALLEGRANZA Oct 04 '23 at 07:23
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2nd question: for the same reason. Apply the rule to $x=0$ as $\phi$ and $\phi$ as only element in $\Gamma$. – Mauro ALLEGRANZA Oct 04 '23 at 07:24
Do the two agree with each other? – Tim Oct 04 '23 at 07:56
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The first one is not a rule: it is an axiom. For Enderton's "equivalent" result wrt Ebbinghaus' rule see Enderton, page 117: GENERALIZATION THEOREM. – Mauro ALLEGRANZA Oct 04 '23 at 08:11
2nd question: if $\Gamma \vdash \phi$, and $x$ is not free in $\Gamma$, then is $x$ not free in $\phi$? – Tim Oct 04 '23 at 08:16
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Because if $x$ is not free in $\phi$ we were unable to "generalize". The issue is: how to "generalize" without committing fallacies. – Mauro ALLEGRANZA Oct 04 '23 at 08:44
See also this post as well as this one. – Mauro ALLEGRANZA Oct 04 '23 at 08:46
if Γ⊢ϕ, and x is not free in Γ, then is x not free in ϕ? – Tim Oct 04 '23 at 09:39
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No. With a $\Gamma$ whatever we have $\Gamma \vdash x=x$ – Mauro ALLEGRANZA Oct 04 '23 at 10:49
Are the two completely unrelated? In the first "α→∀xα if x does not occur free in α", if x is not free in $\alpha$, then adding ∀x to α doesn't matter, correct? – Tim Oct 06 '23 at 06:35

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