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I Swear It Was Here A Minute Ago

Term Neglect

General Neglect

Let's say you have a binary truth function, $(X,Y)_\alpha$, and you want to see if the function when $X=0$ is the same as the function when $X=1$. Naturally, that would be done by saying:

$$(0,Y)_\alpha=(1,Y)_\alpha$$

But, what about writing it completely as a generalized truth function. In that case, the equality would be a biconditional, and rather than entering zero and one for $X$, each function would have a condition value of $(X)$ set to zero or one.

$$\big((X,Y)_\alpha\big|_{(X)_\text{CV}=0},(X,Y)_\alpha\big|_{(X)_\text{CV}=1}\big)_\mathtt{0x9}$$

This is a neglect function.

A neglect function is a special kind of truth function. In neglect, some function is evaluated at all condition values of some domain, and the resulting functions or parses are used as the domain of the neglect function. The function bing evaluated is called the neglected function, and the domain from which the condition values are derived is call the neglected domain. In the example above, the neglected function is $(X,Y)_\alpha$, the neglected domain is $(X)$, and the neglect function's operator is $\mathtt{0x9}$.

As you might infer, neglect functions can get cumbersome when written this way, especially when the neglected domain gets larger. To that end, I have notation that signifies neglect.

$$\big((\textbf{A}^n,\textbf{B}^m)_\alpha\big|_{(\textbf{A}^n)_\text{CV}=0},\ldots,(\textbf{A}^n,\textbf{B}^m)_\alpha\big|_{(\textbf{A}^n)_\text{CV}=2^n-1})_\phi=\big<(\textbf{A}^n,\textbf{B}^m)_\alpha\big>^{(\textbf{A}^n)}_\phi$$

In this notation the neglected function, $(\textbf{A}^n,\textbf{B}^m)_\alpha$, is enclosed in angle brackets, the neglected domain, $(\textbf{A}^n)$, is a superscript to the right angle bracket, and the neglect function's operator, $\phi$, is a subscript to the right angle bracket.

Conjunctive and Disjunctive Neglect

So far in the application of this notation, I have come across two places where neglect can be useful: quantifiers and set operations. To support those applications we will use two very specific forms of neglect. These forms are conjunctive and disjunctive neglect.

Conjunctive neglect requires that the neglect function return one for all values of the neglected domain. As the name might suggest, this is done with repeated conjunction.

$$\Bigg<\bigg<\ldots\Big<\big<(\textbf{A}^n,\textbf{B}^m)_\alpha\big>^{(A_0)}_\mathtt{0x8}\Big>^{(A_1)}_\mathtt{0x8}\ldots\bigg>^{(A_{n-2})}_\mathtt{0x8}\Bigg>^{(A_{n-1})}_\mathtt{0x8}$$

Since conjunctive neglect is so special, so cumbersome to write, and unrestricted in the size of the neglected domain, I have a special notation just for it. This notation takes the form of normal neglect, but rather than a value or variable for the neglect operator, I write the set intersection symbol ($\cap$).

$$\Bigg<\bigg<\ldots\Big<\big<(\textbf{A}^n,\textbf{B}^m)_\alpha\big>^{(A_0)}_\mathtt{0x8}\Big>^{(A_1)}_\mathtt{0x8}\ldots\bigg>^{(A_{n-2})}_\mathtt{0x8}\Bigg>^{(A_{n-1})}_\mathtt{0x8}=\big<(\textbf{A}^n,\textbf{B}^m)_\alpha\big>^{(\textbf{A}^n)}_\cap$$

As you likely suspect, disjunctive neglect returns one if any of the neglected domain's condition values causes the function to return one. This is done by repeated disjunction.

$$\Bigg<\bigg<\ldots\Big<\big<(\textbf{A}^n,\textbf{B}^m)_\alpha\big>^{(A_0)}_\mathtt{0xE}\Big>^{(A_1)}_\mathtt{0xE}\ldots\bigg>^{(A_{n-2})}_\mathtt{0xE}\Bigg>^{(A_{n-1})}_\mathtt{0xE}$$

Again, the value and mass of this form of neglect warrants its own notation. Like its conjunctive cousin, disjunctive neglect is written like normal neglect, but this time the neglect operator is a set union symbol ($\cup$).

$$\Bigg<\bigg<\ldots\Big<\big<(\textbf{A}^n,\textbf{B}^m)_\alpha\big>^{(A_0)}_\mathtt{0xE}\Big>^{(A_1)}_\mathtt{0xE}\ldots\bigg>^{(A_{n-2})}_\mathtt{0xE}\Bigg>^{(A_{n-1})}_\mathtt{0xE}=\big<(\textbf{A}^n,\textbf{B}^m)_\alpha\big>^{(\textbf{A}^n)}_\cup$$

Commutativity of Conjunctive and Disjunctive Neglect

As mentioned, conjunctive and disjunctive neglect are based on repeated conjunction and disjunction. Given that conjunction and disjunction are both self-commutative and self-associative, it can be easily shown that conjunctive and disjunctive neglect are commutative.

First we'll start with a general conjunctive neglect.

$$\big<(\textbf{A}^n,\textbf{B}^m)_\alpha\big>^{(\textbf{A}^n)}_\cap=\Bigg<\bigg<\ldots\Big<\big<(\textbf{A}^n,\textbf{B}^m)_\alpha\big>^{(A_0)}_\mathtt{0x8}\Big>^{(A_1)}_\mathtt{0x8}\ldots\bigg>^{(A_{n-2})}_\mathtt{0x8}\Bigg>^{(A_{n-1})}_\mathtt{0x8}=\big<(\textbf{A}^n,\textbf{B}^m)_\alpha\big>^{(\textbf{A}^n)}_\cap$$

For the sake of simplicity, let's write the inner $n-2$ neglect functions as the truth function they would be, after they were evaluated.

$$\Big<\big<(A_{n-2},A_{n-1},\textbf{B}^m)_\beta\big>^{(A_{n-2})}_\mathtt{0x8}\Big>^{(A_{n-1})}_\mathtt{0x8}$$

Then we'll expand the outer two neglect functions, to their truth function form.

$\Big(\big((A_{n-2},A_{n-1},\textbf{B}^m)_\beta\big|_{A_{n-2}=0}\Big|_{A_{n-1}=0},(A_{n-2},A_{n-1},\textbf{B}^m)_\beta\big|_{A_{n-2}=1}\Big|_{A_{n-1}=0}\big)_\mathtt{0x8},$

$,\big((A_{n-2},A_{n-1},\textbf{B}^m)_\beta\big|_{A_{n-2}=0}\Big|_{A_{n-1}=1},(A_{n-2},A_{n-1},\textbf{B}^m)_\beta\big|_{A_{n-2}=1}\Big|_{A_{n-1}=1}\big)_\mathtt{0x8}\Big)_\mathtt{0x8}$

Again, let's pair this down a little.

$$\Big(\big((\textbf{B}^m)_{\beta_0},(\textbf{B}^m)_{\beta_1}\big)_\mathtt{0x8},\big((\textbf{B}^m)_{\beta_2},(\textbf{B}^m)_{\beta_3}\big)_\mathtt{0x8}\Big)_\mathtt{0x8}$$

Now, for the algebra. By associativity we can group $\beta_1$ with the conjunction of $\beta_2$ and $\beta_3$, leaving $\beta_0$ alone.

$$\bigg((\textbf{B}^m)_{\beta_0},\Big((\textbf{B}^m)_{\beta_1},\big((\textbf{B}^m)_{\beta_2},(\textbf{B}^m)_{\beta_3}\big)_\mathtt{0x8}\Big)_\mathtt{0x8}\bigg)_\mathtt{0x8}$$

By commutativity, $\beta_2$ and $\beta_3$ can swap places.

$$\bigg((\textbf{B}^m)_{\beta_0},\Big((\textbf{B}^m)_{\beta_1},\big((\textbf{B}^m)_{\beta_3},(\textbf{B}^m)_{\beta_2}\big)_\mathtt{0x8}\Big)_\mathtt{0x8}\bigg)_\mathtt{0x8}$$

Again by associativity, $\beta_1$ and $\beta_3$ can be grouped together, leaving $\beta_2$ alone.

$$\bigg((\textbf{B}^m)_{\beta_0},\Big(\big((\textbf{B}^m)_{\beta_1},(\textbf{B}^m)_{\beta_3}\big)_\mathtt{0x8},(\textbf{B}^m)_{\beta_2}\Big)_\mathtt{0x8}\bigg)_\mathtt{0x8}$$

A little more commutativity swaps $\beta_2$ with the conjunction of $\beta_1$ and $\beta_3$.

$$\bigg((\textbf{B}^m)_{\beta_0},\Big((\textbf{B}^m)_{\beta_2},\big((\textbf{B}^m)_{\beta_1},(\textbf{B}^m)_{\beta_3}\big)_\mathtt{0x8}\Big)_\mathtt{0x8}\bigg)_\mathtt{0x8}$$

Then, by associativity, $\beta_0$ can be grouped with $\beta_2$, while leaving the conjunction between $\beta_1$ and $\beta_3$ alone.

$$\Big(\big((\textbf{B}^m)_{\beta_0},(\textbf{B}^m)_{\beta_2}\big),\big((\textbf{B}^m)_{\beta_1},(\textbf{B}^m)_{\beta_3}\big)_\mathtt{0x8}\Big)_\mathtt{0x8}$$

By back tracking the substitutions made earlier, we can see that this function corresponds to neglecting $A_{2^n-1}$ before $A_{2^n-2}$. Conjunctive neglect is commutative.

$$\Big<\big<(A_{n-2},A_{n-1},\textbf{B}^m)_\beta\big>^{(A_{n-2})}_\mathtt{0x8}\Big>^{(A_{n-1})}_\mathtt{0x8}=\Big<\big<(A_{n-2},A_{n-1},\textbf{B}^m)_\beta\big>^{(A_{n-1})}_\mathtt{0x8}\Big>^{(A_{n-2})}_\mathtt{0x8}$$

The proof for the commutativity of disjunctive neglect is identical.

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