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Revision as of 19:08, 23 January 2009
, 19:08, 23 January 2009→Stretching Exercises: mathematical markup
: <math>F(x, y) ~=~ F_{6}^{(2)} (x, y) ~=~ \underline{(}~x~,~y~\underline{)}</math>
: <math>F(x, y) ~=~ F_{6}^{(2)} (x, y) ~=~ \underline{(}~x~,~y~\underline{)}</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\!</math> in <math>\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>
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:
<pre>
<pre>
[| -[S]- |] = [| F |]
[| -[S]- |] = [| F |]
