User:Jon Awbrey/Cactus Language Stretching Exercises - Revision history

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@@ Line 24: / Line 24: @@
 [| \downharpoonleft s \downharpoonright |]
 & = & [| F |]
-\\[6pt]
+\\[4pt]
 & = & F^{-1} (1)
-\\[6pt]
+\\[4pt]
 & = & \{~ (x, y) \in \mathbb{B}^2 ~:~ s ~\}
-\\[6pt]
+\\[4pt]
 & = & \{~ (x, y) \in \mathbb{B}^2 ~:~ F(x, y) = 1 ~\}
-\\[6pt]
+\\[4pt]
 & = & \{~ (x, y) \in \mathbb{B}^2 ~:~ F(x, y) ~\}
-\\[6pt]
+\\[4pt]
 & = & \{~ (x, y) \in \mathbb{B}^2 ~:~ \texttt{(}~x~,~y~\texttt{)} = 1 ~\}
-\\[6pt]
+\\[4pt]
 & = & \{~ (x, y) \in \mathbb{B}^2 ~:~ \texttt{(}~x~,~y~\texttt{)} ~\}
-\\[6pt]
+\\[4pt]
 & = & \{~ (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 ~\}
-\\[6pt]
+\\[4pt]
 & = & \{~ (x, y) \in \mathbb{B}^2 ~:~ x ~\mathrm{not~equal~to}~ y ~\}
-\\[6pt]
+\\[4pt]
 & = & \{~ (x, y) \in \mathbb{B}^2 ~:~ x \nLeftrightarrow y ~\}
-\\[6pt]
+\\[4pt]
 & = & \{~ (x, y) \in \mathbb{B}^2 ~:~ x \neq y ~\}
-\\[6pt]
+\\[4pt]
 & = & \{~ (x, y) \in \mathbb{B}^2 ~:~ x + y ~\}.
 \end{array}</math>
@@ Line 62: / Line 62: @@
+|
 <math>\begin{matrix}
 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
 \end{matrix}</math>
 |}
@@ Line 90: / Line 75: @@
+|
 <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}
 \end{array}</math>
 |}
@@ Line 112: / Line 87: @@
 <math>\begin{matrix}
 F^\$ (p, q)(x) & = & \underline{(}~p~,~q~\underline{)}^\$ (x) & \in & \mathbb{B}
-\\
+\\[4pt]
 \Updownarrow   &   & \Updownarrow
-\\
+\\[4pt]
 F(p(x), q(x))  & = & \underline{(}~p(x)~,~q(x)~\underline{)}  & \in & \mathbb{B}
 \end{matrix}</math>
 |}
@@ Line 129: / Line 101: @@
 [| F^\$ (p, q) |]
 & = & [| \underline{(}~p~,~q~\underline{)}^\$ |]
-\\[6pt]
+\\[4pt]
 & = & (F^\$ (p, q))^{-1} (1)
-\\[6pt]
+\\[4pt]
 & = & \{~ x \in X ~:~ F^\$ (p, q)(x) ~\}
-\\[6pt]
+\\[4pt]
 & = & \{~ x \in X ~:~ \underline{(}~p~,~q~\underline{)}^\$ (x) ~\}
-\\[6pt]
+\\[4pt]
 & = & \{~ x \in X ~:~ \underline{(}~p(x)~,~q(x)~\underline{)} ~\}
-\\[6pt]
+\\[4pt]
 & = & \{~ x \in X ~:~ p(x) + q(x) ~\}
-\\[6pt]
+\\[4pt]
 & = & \{~ x \in X ~:~ p(x) \neq q(x) ~\}
-\\[6pt]
+\\[4pt]
 & = & \{~ x \in X ~:~ \upharpoonleft P \upharpoonright (x) ~\neq~ \upharpoonleft Q \upharpoonright (x) ~\}
-\\[6pt]
+\\[4pt]
 & = & \{~ 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 ~\}
-\\[6pt]
+\\[4pt]
 & = & \{~ x \in X ~:~ x \in P\!-\!Q ~\cup~ Q\!-\!P ~\}
-\\[6pt]
+\\[4pt]
 & = & \{~ x \in X ~:~ x \in P + Q ~\}
-\\[6pt]
+\\[4pt]
 & = & P + Q ~\subseteq~ X
-\\[6pt]
+\\[4pt]
 & = & [|p|] + [|q|] ~\subseteq~ X
 \end{array}</math>
 |}

@@ Line 9: / Line 9: @@
 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>
+{| align="center" cellpadding="8"
 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}
@@ Line 55: / Line 57: @@
 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}
@@ Line 83: / Line 85: @@
 |}
-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}
@@ Line 104: / Line 106: @@
 |}
-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}
@@ Line 120: / Line 122: @@
 |}
-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}