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→‎Stretching Exercises: mathematical markup
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The connection in question is a boolean function on the variables <math>x, y\!</math> that returns a value of <math>\underline{1}</math> just when just one of the pair <math>x, y\!</math> is not equal to <math>\underline{1},</math> or what amounts to the same thing, just when just one of the pair <math>x, y\!</math> is equal to <math>\underline{1}.</math>  There is clearly an isomorphism between this connection, viewed as an operation on the boolean domain <math>\underline\mathbb{B} = \{ \underline{0}, \underline{1} \},</math> and the dyadic operation on binary values <math>x, y \in \mathbb{B} = \operatorname{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>\underline{1}</math> just when just one of the pair <math>x, y\!</math> is not equal to <math>\underline{1},</math> or what amounts to the same thing, just when just one of the pair <math>x, y\!</math> is equal to <math>\underline{1}.</math>  There is clearly an isomorphism between this connection, viewed as an operation on the boolean domain <math>\underline\mathbb{B} = \{ \underline{0}, \underline{1} \},</math> and the dyadic operation on binary values <math>x, y \in \mathbb{B} = \operatorname{GF}(2)</math> that is otherwise known as <math>x + y\!.</math>
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The same connection <math>F : \underline\mathbb{B}^2 \to \underline\mathbb{B}</math> can also be read as a proposition about things in the universe <math>X = \underline\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>^{\backprime\backprime} ~x~ \operatorname{is~not~equal~to} ~y~ ^{\prime\prime}.</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:
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The same connection <math>F : \underline\mathbb{B}^2 \to \underline\mathbb{B}</math> can also be read as a proposition about things in the universe <math>X = \underline\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>^{\backprime\backprime} \, x ~\operatorname{is~not~equal~to}~ y \, ^{\prime\prime}.</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:
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{| align="center" cellpadding="8" width="90%"
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|
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<math>\begin{array}{lll}
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[| \downharpoonright s \downharpoonleft |]
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& = & [| F |]
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\\
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\\
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& = & F^{-1} (\underline{1})
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\\
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\\
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& = & \{~ (x, y) \in \underline\mathbb{B}^2 ~:~ s ~\}
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\\
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\\
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& = & \{~ (x, y) \in \underline\mathbb{B}^2 ~:~ F(x, y) = \underline{1} ~\}
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\\
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\\
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& = & \{~ (x, y) \in \underline\mathbb{B}^2 ~:~ F(x, y) ~\}
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\\
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\\
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& = & \{~ (x, y) \in \underline\mathbb{B}^2 ~:~ \underline{(}~x~,~y~\underline{)} = \underline{1} ~\}
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\\
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\\
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& = & \{~ (x, y) \in \underline\mathbb{B}^2 ~:~ \underline{(}~x~,~y~\underline{)} ~\}
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\\
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\\
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& = & \{~ (x, y) \in \underline\mathbb{B}^2 ~:~ x ~\operatorname{exclusive~or}~ y ~\}
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\\
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\\
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& = & \{~ (x, y) \in \underline\mathbb{B}^2 ~:~ \operatorname{just~one~true~of}~ x, y ~\}
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\\
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\\
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& = & \{~ (x, y) \in \underline\mathbb{B}^2 ~:~ x ~\operatorname{not~equal~to}~ y ~\}
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\\
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\\
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& = & \{~ (x, y) \in \underline\mathbb{B}^2 ~:~ x \nLeftrightarrow y ~\}
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\\
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\\
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& = & \{~ (x, y) \in \underline\mathbb{B}^2 ~:~ x \neq y ~\}
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\\
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\\
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& = & \{~ (x, y) \in \underline\mathbb{B}^2 ~:~ x + y ~\}.
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\end{array}</math>
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|}
    
<pre>
 
<pre>
[| -[S]- |]  =  [| F |]
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            =  F^(-1)(%1%)
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            =  {<x, y> in %B%^2  :  S}
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            =  {<x, y> in %B%^2  :  F(x, y) = %1%}
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            =  {<x, y> in %B%^2  :  F(x, y)}
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            =  {<x, y> in %B%^2  :  -(x, y)- = %1%}
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            =  {<x, y> in %B%^2  :  -(x, y)- }
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            =  {<x, y> in %B%^2  :  x exclusive-or y}
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            =  {<x, y> in %B%^2  :  just one true of x, y}
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            =  {<x, y> in %B%^2  :  x not equal to y}
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            =  {<x, y> in %B%^2  :  x <=/=> y}
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            =  {<x, y> in %B%^2  :  x =/= y}
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            =  {<x, y> in %B%^2  :  x + y}.
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Notice the slight distinction, that I continue to maintain at this point,
 
Notice the slight distinction, that I continue to maintain at this point,
 
between the logical values {false, true} and the algebraic values {0, 1}.
 
between the logical values {false, true} and the algebraic values {0, 1}.
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