Changes

When the cactus notation is used to represent boolean functions, a single <math>\$</math> sign at the end of the expression is enough to remind the reader that the connections are meant to be stretched to several propositions on a universe <math>X.</math>

When the cactus notation is used to represent boolean functions, a single <math>\$</math> sign at the end of the expression is enough to remind the reader that the connections are meant to be stretched to several propositions on a universe <math>X.</math>

For example, take the connection <math>F : \mathbb{B}^2 \to \mathbb{B}</math> such that:

For example, take the connection <math>F : \mathbb{B}^2 \to \mathbb{B}</math> defined as follows.

: <math>F(x, y) ~=~ F_{6}^{(2)} (x, y) ~=~ \texttt{(}~x~,~y~\texttt{)}</math>

| <math>F(x, y) ~=~ F_{6}^{(2)} (x, y) ~=~ \texttt{(}~x~,~y~\texttt{)}</math>

The connection in question is a boolean function on the variables <math>x, y</math> that returns a value of <math>1</math> just when just one of the pair <math>x, y</math> is not equal to <math>1,</math> or what amounts to the same thing, just when just one of the pair <math>x, y</math> is equal to <math>1.</math>  There is clearly an isomorphism between this connection, viewed as an operation on the boolean domain <math>\mathbb{B} = \{ 0, 1 \},</math> and the dyadic operation on binary values <math>x, y \in \mathbb{B} = \mathrm{GF}(2)</math> that is otherwise known as <math>x + y.</math>

The connection in question is a boolean function on the variables <math>x, y</math> that returns a value of <math>1</math> just when just one of the pair <math>x, y</math> is not equal to <math>1,</math> or what amounts to the same thing, just when just one of the pair <math>x, y</math> is equal to <math>1.</math>  There is clearly an isomorphism between this connection, viewed as an operation on the boolean domain <math>\mathbb{B} = \{ 0, 1 \},</math> and the dyadic operation on binary values <math>x, y \in \mathbb{B} = \mathrm{GF}(2)</math> that is otherwise known as <math>x + y.</math>

The same connection <math>F : \mathbb{B}^2 \to \mathbb{B}</math> can also be read as a proposition about things in the universe <math>X = \mathbb{B}^2.</math>  If <math>s</math> is a sentence that denotes the proposition <math>F,</math> then the corresponding assertion says exactly what one states in uttering the sentence <math>\text{“} \, x ~\mathrm{is~not~equal~to}~ y \, \text{”}.</math>  In such a case, one has <math>\downharpoonleft s \downharpoonright \, = F,</math> and all of the following expressions are ordinarily taken as equivalent descriptions of the same set:

The same connection <math>F : \mathbb{B}^2 \to \mathbb{B}</math> can also be read as a proposition about things in the universe <math>X = \mathbb{B}^2.</math>  If <math>s</math> is a sentence that denotes the proposition <math>F,</math> then the corresponding assertion says exactly what one states in uttering the sentence <math>\text{“} \, x ~\mathrm{is~not~equal~to}~ y \, \text{”}.</math>  In such a case, one has <math>\downharpoonleft s \downharpoonright \, = F,</math> and all of the following expressions are ordinarily taken as equivalent descriptions of the same set.

{| align="center" cellpadding="8" width="90%"

{| align="center" cellpadding="8"

|

|

<math>\begin{array}{lll}

<math>\begin{array}{lll}

[| \downharpoonleft s \downharpoonright |]

[| \downharpoonleft s \downharpoonright |]

& = & [| F |]

& = & [| F |]

\\[6pt]

\\[4pt]

& = & F^{-1} (1)

& = & F^{-1} (1)

\\[6pt]

\\[4pt]

& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ s ~\}

& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ s ~\}

\\[6pt]

\\[4pt]

& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ F(x, y) = 1 ~\}

& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ F(x, y) = 1 ~\}

\\[6pt]

\\[4pt]

& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ F(x, y) ~\}

& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ F(x, y) ~\}

\\[6pt]

\\[4pt]

& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ \texttt{(}~x~,~y~\texttt{)} = 1 ~\}

& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ \texttt{(}~x~,~y~\texttt{)} = 1 ~\}

\\[6pt]

\\[4pt]

& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ \texttt{(}~x~,~y~\texttt{)} ~\}

& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ \texttt{(}~x~,~y~\texttt{)} ~\}

\\[6pt]

\\[4pt]

& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ x ~\mathrm{exclusive~or}~ y ~\}

& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ x ~\mathrm{exclusive~or}~ y ~\}

\\[6pt]

\\[4pt]

& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ \mathrm{just~one~true~of}~ x, y ~\}

& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ \mathrm{just~one~true~of}~ x, y ~\}

\\[6pt]

\\[4pt]

& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ x ~\mathrm{not~equal~to}~ y ~\}

& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ x ~\mathrm{not~equal~to}~ y ~\}

\\[6pt]

\\[4pt]

& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ x \nLeftrightarrow y ~\}

& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ x \nLeftrightarrow y ~\}

\\[6pt]

\\[4pt]

& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ x \neq y ~\}

& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ x \neq y ~\}

\\[6pt]

\\[4pt]

& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ x + y ~\}.

& = & \{~ (x, y) \in \mathbb{B}^2 ~:~ x + y ~\}.

\end{array}</math>

\end{array}</math>

This covers the properties of the connection <math>F(x, y) = \underline{(}~x~,~y~\underline{)},</math> treated as a proposition about things in the universe <math>X = \mathbb{B}^2.</math>  Staying with this same connection, it is time to demonstrate how it can be "stretched" to form an operator on arbitrary propositions.

This covers the properties of the connection <math>F(x, y) = \underline{(}~x~,~y~\underline{)},</math> treated as a proposition about things in the universe <math>X = \mathbb{B}^2.</math>  Staying with this same connection, it is time to demonstrate how it can be "stretched" to form an operator on arbitrary propositions.

To continue the exercise, let <math>p</math> and <math>q</math> be arbitrary propositions about things in the universe <math>X,</math> that is, maps of the form <math>p, q : X \to \mathbb{B},</math> and suppose that <math>p, q</math> are indicator functions of the sets <math>P, Q \subseteq X,</math> respectively. In other words, we have the following data:

To continue the exercise, let <math>p</math> and <math>q</math> be arbitrary propositions about things in the universe <math>X,</math> that is, maps of the form <math>p, q : X \to \mathbb{B},</math> and suppose that <math>p, q</math> are indicator functions of the sets <math>P, Q \subseteq X,</math> respectively.&nbsp; In other words, we have the following data.

{| align="center" cellpadding="8" width="90%"

{| align="center" cellpadding="8"

|

|

<math>\begin{matrix}

<math>\begin{matrix}

p

p & = & \upharpoonleft P \upharpoonright & : & X \to \mathbb{B}

& = &

\\[4pt]

\upharpoonleft P \upharpoonright

q & = & \upharpoonleft Q \upharpoonright & : & X \to \mathbb{B}

& : &

\\[4pt]

X \to \mathbb{B}

(p, q) & = & (\upharpoonleft P \upharpoonright, \upharpoonleft Q \upharpoonright) & : & (X \to \mathbb{B})^2

\\

q

& = &

\upharpoonleft Q \upharpoonright

& : &

X \to \mathbb{B}

\\

(p, q)

& = &

(\upharpoonleft P \upharpoonright, \upharpoonleft Q \upharpoonright)

& : &

(X \to \mathbb{B})^2

\end{matrix}</math>

\end{matrix}</math>

|}

|}

Then one has an operator <math>F^\$,</math> the stretch of the connection <math>F</math> over <math>X,</math> and a proposition <math>F^\$ (p, q),</math> the stretch of <math>F</math> to <math>(p, q)</math> on <math>X,</math> with the following properties:

Then one has an operator <math>F^\$,</math> the stretch of the connection <math>F</math> over <math>X,</math> and a proposition <math>F^\$ (p, q),</math> the stretch of <math>F</math> to <math>(p, q)</math> on <math>X,</math> with the following properties.

{| align="center" cellpadding="8" width="90%"

{| align="center" cellpadding="8"

|

|

<math>\begin{array}{ccccl}

<math>\begin{array}{ccccl}

F^\$

F^\$ & = & \underline{(} \ldots, \ldots \underline{)}^\$ & : & (X \to \mathbb{B})^2 \to (X \to \mathbb{B})

& = &

\\[4pt]

\underline{(} \ldots, \ldots \underline{)}^\$

F^\$ (p, q) & = & \underline{(}~p~,~q~\underline{)}^\$ & : & X \to \mathbb{B}

& : &

(X \to \mathbb{B})^2 \to (X \to \mathbb{B})

\\

F^\$ (p, q)

& = &

\underline{(}~p~,~q~\underline{)}^\$

& : &

X \to \mathbb{B}

\end{array}</math>

\end{array}</math>

|}

|}

As a result, the application of the proposition <math>F^\$ (p, q)</math> to each <math>x \in X</math> returns a logical value in <math>\mathbb{B},</math> all in accord with the following equations:

As a result, the application of the proposition <math>F^\$ (p, q)</math> to each <math>x \in X</math> returns a logical value in <math>\mathbb{B},</math> all in accord with the following equations.

{| align="center" cellpadding="8" width="90%"

{| align="center" cellpadding="8"

|

|

<math>\begin{matrix}

<math>\begin{matrix}

F^\$ (p, q)(x) & = & \underline{(}~p~,~q~\underline{)}^\$ (x) & \in & \mathbb{B}

F^\$ (p, q)(x) & = & \underline{(}~p~,~q~\underline{)}^\$ (x) & \in & \mathbb{B}

\\

\\[4pt]

\Updownarrow  &  & \Updownarrow

\Updownarrow  &  & \Updownarrow

\\

\\[4pt]

F(p(x), q(x))  & = & \underline{(}~p(x)~,~q(x)~\underline{)}  & \in & \mathbb{B}

F(p(x), q(x))  & = & \underline{(}~p(x)~,~q(x)~\underline{)}  & \in & \mathbb{B}

\end{matrix}</math>

\end{matrix}</math>

|}

|}

For each choice of propositions <math>p</math> and <math>q</math> about things in <math>X,</math> the stretch of <math>F</math> to <math>p</math> and <math>q</math> on <math>X</math> is just another proposition about things in <math>X,</math> a simple proposition in its own right, no matter how complex its current expression or its present construction as <math>F^\$ (p, q) = \underline{(}~p~,~q~\underline{)}^\$</math> makes it appear in relation to <math>p</math> and <math>q.</math>  Like any other proposition about things in <math>X,</math> it indicates a subset of <math>X,</math> namely, the fiber that is variously described in the following ways:

For each choice of propositions <math>p</math> and <math>q</math> about things in <math>X,</math> the stretch of <math>F</math> to <math>p</math> and <math>q</math> on <math>X</math> is just another proposition about things in <math>X,</math> a simple proposition in its own right, no matter how complex its current expression or its present construction as <math>F^\$ (p, q) = \underline{(}~p~,~q~\underline{)}^\$</math> makes it appear in relation to <math>p</math> and <math>q.</math>  Like any other proposition about things in <math>X,</math> it indicates a subset of <math>X,</math> namely, the fiber that is variously described in the following ways.

{| align="center" cellpadding="8" width="90%"

{| align="center" cellpadding="8"

|

|

<math>\begin{array}{lll}

<math>\begin{array}{lll}

[| F^\$ (p, q) |]

[| F^\$ (p, q) |]

& = & [| \underline{(}~p~,~q~\underline{)}^\$ |]

& = & [| \underline{(}~p~,~q~\underline{)}^\$ |]

\\[6pt]

\\[4pt]

& = & (F^\$ (p, q))^{-1} (1)

& = & (F^\$ (p, q))^{-1} (1)

\\[6pt]

\\[4pt]

& = & \{~ x \in X ~:~ F^\$ (p, q)(x) ~\}

& = & \{~ x \in X ~:~ F^\$ (p, q)(x) ~\}

\\[6pt]

\\[4pt]

& = & \{~ x \in X ~:~ \underline{(}~p~,~q~\underline{)}^\$ (x) ~\}

& = & \{~ x \in X ~:~ \underline{(}~p~,~q~\underline{)}^\$ (x) ~\}

\\[6pt]

\\[4pt]

& = & \{~ x \in X ~:~ \underline{(}~p(x)~,~q(x)~\underline{)} ~\}

& = & \{~ x \in X ~:~ \underline{(}~p(x)~,~q(x)~\underline{)} ~\}

\\[6pt]

\\[4pt]

& = & \{~ x \in X ~:~ p(x) + q(x) ~\}

& = & \{~ x \in X ~:~ p(x) + q(x) ~\}

\\[6pt]

\\[4pt]

& = & \{~ x \in X ~:~ p(x) \neq q(x) ~\}

& = & \{~ x \in X ~:~ p(x) \neq q(x) ~\}

\\[6pt]

\\[4pt]

& = & \{~ x \in X ~:~ \upharpoonleft P \upharpoonright (x) ~\neq~ \upharpoonleft Q \upharpoonright (x) ~\}

& = & \{~ x \in X ~:~ \upharpoonleft P \upharpoonright (x) ~\neq~ \upharpoonleft Q \upharpoonright (x) ~\}

\\[6pt]

\\[4pt]

& = & \{~ x \in X ~:~ x \in P ~\nLeftrightarrow~ x \in Q ~\}

& = & \{~ x \in X ~:~ x \in P ~\nLeftrightarrow~ x \in Q ~\}

\\[6pt]

\\[4pt]

& = & \{~ x \in X ~:~ x \in P\!-\!Q ~\mathrm{or}~ x \in Q\!-\!P ~\}

& = & \{~ x \in X ~:~ x \in P\!-\!Q ~\mathrm{or}~ x \in Q\!-\!P ~\}

\\[6pt]

\\[4pt]

& = & \{~ x \in X ~:~ x \in P\!-\!Q ~\cup~ Q\!-\!P ~\}

& = & \{~ x \in X ~:~ x \in P\!-\!Q ~\cup~ Q\!-\!P ~\}

\\[6pt]

\\[4pt]

& = & \{~ x \in X ~:~ x \in P + Q ~\}

& = & \{~ x \in X ~:~ x \in P + Q ~\}

\\[6pt]

\\[4pt]

& = & P + Q ~\subseteq~ X

& = & P + Q ~\subseteq~ X

\\[6pt]

\\[4pt]

& = & [|p|] + [|q|] ~\subseteq~ X

& = & [|p|] + [|q|] ~\subseteq~ X

\end{array}</math>

\end{array}</math>

|}

|}