0

Universal Quantifier

$$\forall x(\mathscr{Y})\Leftrightarrow\Big<\big((\textbf{A}^n)_\alpha,\mathscr{Y}\big)_\mathtt{0xD}\Big>^{(\textbf{A}^n)}_\cap\ni x=(\textbf{A}^n)_\text{CV}$$

Let's start by exploring what the universal quantifier says. Assuming the possible values of $x$ are limited to some set, $\mathbb{X}$, the universal quantifier says that every $x\in\mathbb{X}$ will cause $\mathscr{Y}$ to be true. That is, if $x\in\mathbb{X}$ then $\mathscr{Y}$.

I would be remiss to stop at simply stating a conditional without substantiating it. In the two cases where $x$ is not a member $\mathbb{X}$ the universal quantifier makes no promises about the state of $\mathscr{Y}$, so $\mathscr{Y}$ can be either true or false. On the other hand, when $x$ IS a member of $\mathbb{X}$ we have to be more careful. If $x\in\mathbb{X}$ and $\mathscr{Y}$ is false, then there is an allowed value of $x$ that fails to satisfy $\mathscr{Y}$, thus contradicting the quantifier. While $x\in\mathbb{X}$ when $\mathscr{Y}$ is true is precisely what a universal quantifier wants.

$$x\in\mathbb{X}\Rightarrow\mathscr{Y}$$

With the target of a universal quantifier reduced to a logical expression, let's work to get the antecedent and the consequent in logical terms. Naturally, $\mathscr{Y}$ is already a predicate, so its logical pedigree is obvious. On the other hand $x\in\mathbb{X}$ requires a little more work. Consider $x$ to be a condition value of some property domain, and $\mathbb{X}$ to be the set of condition values that satisfy some truth function $\alpha$. $$x=(\textbf{A}^n)_\text{CV}\text{ and }\mathbb{X}=\{x\big|(\textbf{A}^n)_\alpha|_{(\textbf{A}^n)_\text{CV}}=1\}$$ Thus, the quantifier is supported when ever $(\textbf{A}^n)_\alpha\Rightarrow\mathscr{Y}$ is true, or using generalized truth functions, when $\big((\textbf{A}^n)_\alpha,\mathscr{Y}\big)_\mathtt{0xD}=1$.

Now we're almost there, but the conditional above can only test one condition value of $\textbf{A}^n$ at a time. In order for the universal quantifier to be fully satisfied, we must test the conditional for every condition value of $\textbf{A}^n$. For this I will introduce a special function, with an unspecified number of input variables, that only returns one if every input variable returns one. This mega-conjunction will be designated as a truth function with the set intersection symbol, $\cap$, in place of the operator. $$\big((\alpha_0,\mathscr{Y}\big)_\mathtt{0xD},\ldots,\big(\alpha_{2^n-1},\mathscr{Y})\big)_\cap$$ Then, since each conjunct above involves a unique condition value of $\textbf{A}^n$, this conjunction can be rewritten as a neglect function, with the same $\cap$ as the neglect operator.

$$\Big<\big((\textbf{A}^n)_\alpha,\mathscr{Y}\big)_\mathtt{0xD}\Big>^{(\textbf{A}^n)}_\cap$$
< Previous
Next >