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Let $B$ be a complete Boolean algebra. Jech defines a Boolean-valued model $\mathfrak{A}$ of the language of set theory to consist of a Boolean universe $A$ and functions of two variables with values in $B$, $\qquad \qquad \| x=y \|, \qquad \| x \in y \|$

that safisfy the following:

$ (i)\ \ \ \| x=x \| = 1 \\ (ii) \ \ \| x=y \| = \|y = x \| \\ (iii)\ \| x=y \| \cdot \| y=z \| \leq \| x=z \| \\ (iv) \ \ \| x\in y \| \cdot \|v=x \| \cdot \|w=y \| \leq \|v \in w \| $

For every formula $\phi(a_1,\ldots,a_n)$, we define the Boolean value $\| \phi(a_1,\ldots,a_n) \|$ of $\phi$ as follows:

$(a)$ For atomic formulas, we have the functions $\| x=y \|, \| x \in y \| $

$(b)$ If $\phi$ is negation, disjunction, $\exists x \, \psi$,

$\qquad \| \lnot \phi(a_1,\ldots,a_n) \| = - \| \phi(a_1,\ldots,a_n) \| $

$\qquad \| (\phi \vee \psi)(a_1,\ldots,a_n) \| = \| \phi(a_1,\ldots,a_n) \| + \| \psi (a_1,\ldots,a_n) \|$

$\qquad \| \exists x \, \psi(x,a_1,\ldots,a_n) \| = \underset{a\in A}{\sum} \| \psi (a,a_1,\ldots,a_n)\|$

Next, Jech says that it's easy to prove

$\qquad \qquad \qquad \| x=y \| \cdot \|\phi(x)\| \leq \|\phi(y)\|$

However, I'm not seeing it. It seems clear that it should be a proof by induction on the complexity of $\phi$, and the inductive step is easy in the case of disjunction and existential, but why does it hold for negation? That is, why is it that if $ \| x=y \| \cdot \|\phi(x)\| \leq \|\phi(y)\|$, then $\| x=y \| \cdot \|\lnot \phi(x)\| \leq \| \lnot \phi(y)\|$?

Thank you!

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    $\begingroup$ I think the easiest way to see this is by noting that by (i)-(iv) all atomic instances of the identity axioms $x= y \to (\phi(x) \leftrightarrow \phi(y))$ get value 1 and then note that the non-atomic instances follow in first-order logic by a simple induction. So those instances too will have value 1. $\endgroup$ Nov 28 '15 at 23:07
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    $\begingroup$ @SamRoberts: or to turn that around very slightly, to make it sound closer to the original approach: instead of directly showing by induction $\| x=y\| \cdot \| \varphi(x) \| \leq \|\varphi(y) \|$, strengthen the statement of the induction to $\| x = y \| \leq \left( \| \varphi(x) \| \leftrightarrow \| \varphi(y) \| \right) $. $\endgroup$ Nov 28 '15 at 23:14
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    $\begingroup$ Or, one can simply use contraposition. Namely, in any Boolean algebra, $a\cdot b\leq c$ just in case $a\cdot\neg c\leq \neg b$. So from $\|y=x\|\cdot\|\varphi(y)\|\leq\|\varphi(x)\|$ and the fact that $\|x=y\|=\|y=x\|$, we deduce $\|x=y\|\cdot\|\neg\varphi(x)\|\leq\|\neg\varphi(y)\|$, as desired. $\endgroup$ Nov 29 '15 at 0:17
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    $\begingroup$ Meanwhile, it is probably better to post answers as answers, since the site works better that way. @SamRoberts, could you post your answer? $\endgroup$ Nov 29 '15 at 0:21
  • $\begingroup$ @JoelDavidHamkins Sure! (I thought it might have been moved to stackexchange.) $\endgroup$ Nov 29 '15 at 0:38
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One way to show this is by first noting that (i)-(iv) ensure the atomic instances of the identity axioms:

$x = y \to [\phi(x) \leftrightarrow \phi(y)]$

get value 1. Then observe, by a simple induction on the complexity of $\phi$, that the non-atomic instances follow from the atomic instances in first-order logic. So they too get value 1.

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