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Existential Quantifier

$$\exists x(\mathscr{Y})\Leftrightarrow\Big<\big((\textbf{A}^n)_\alpha,\mathscr{Y}\big)_\mathtt{0x8}\Big>^{(\textbf{A}^n)}_\cup\ni x=(\textbf{A}^n)_\text{CV}$$

We'll start by looking into what the existential quantifier means. Assuming the possible values of $x$ are restricted to some set, $\mathbb{X}$, the existential quantifier says that there is at least one value of $x$ where $x\in\mathbb{X}$ and $\mathscr{Y}$ are both true.

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

With the purpose of the existential quantifier reduced to a logical expression, let's work on getting the conjuncts in logical terms. Since $\mathscr{Y}$ is a predicate it is already in logical terms. However, $x\in\mathbb{X}$ is a little more complicated. 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 $\alpha$. $$x=(\textbf{A}^n)_\text{CV}\text{ and }\mathbb{X}=\big\{x\big|(\textbf{A}^n)_\alpha|_{(\textbf{A}^n)_\text{CV}}=1\big\}$$ Thus, the quantifier is supported whenever $(\textbf{A}^n)_\alpha\land\mathscr{Y}$ is true, or using generalized truth functions, when $\big((\textbf{A}^n)_\alpha,\mathscr{Y}\big)_\mathtt{0x8}=1$.

Now, the conjunction above can only test one condition value of $\textbf{A}^n$ at a time. In order fro the existential quantifier to be fully satisfied, we must test the conjunction for every condition value of $\textbf{A}^n$. To test these condition values we will need a neglect function that returns one if at least one condition value of $\textbf{A}^n$ satisfies the conjunction. Since this function most resembles a large disjunction, and disjunctions are similar to set unions, I will put at $\cup$ symbol in place of the neglect operator. Thus, $$\Big<\big((\textbf{A}^n)_\alpha,\mathscr{Y}\big)_\mathtt{0x8}\Big>^{(\textbf{A}^n)}_\cup$$

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