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Revision as of 14:52, 9 March 2009
, 14:52, 9 March 2009→Note 10: texttt
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<math>\begin{array}{lccccc}
<math>\begin{array}{lccccc}
−\text{Alias Map.} & (x, y) & = & F(u, v) & = & ( ~\underline{((}~ u ~\underline{)(}~ v ~\underline{))}~ , ~\underline{((}~ u ~,~ v ~\underline{))}~ )
+\text{Alias Map.} & (x, y) & = & F(u, v) & = & ( ~\texttt{((u)(v))}~ , ~\texttt{((u,~v))}~ )
\\ \\
\\ \\
−\text{Alibi Map.} & (u', v') & = & F(u, v) & = & ( ~\underline{((}~ u ~\underline{)(}~ v ~\underline{))}~ , ~\underline{((}~ u ~,~ v ~\underline{))}~ )
+\text{Alibi Map.} & (u', v') & = & F(u, v) & = & ( ~\texttt{((u)(v))}~ , ~\texttt{((u,~v))}~ )
\end{array}</math>
\end{array}</math>
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Maintaining a strict analogy with ordinary difference calculus would perhaps have us write <math>\operatorname{D}G_j = \operatorname{E}G_j - G_j,</math> but the sum and difference operations are the same thing in boolean arithmetic. It is more often natural in the logical context to consider an initial proposition <math>q,\!</math> then to compute the enlargement <math>\operatorname{E}q,</math> and finally to determine the difference <math>\operatorname{D}q = q + \operatorname{E}q,</math> so we let the variant order of terms reflect this sequence of considerations.
Maintaining a strict analogy with ordinary difference calculus would perhaps have us write <math>\operatorname{D}G_j = \operatorname{E}G_j - G_j,</math> but the sum and difference operations are the same thing in boolean arithmetic. It is more often natural in the logical context to consider an initial proposition <math>q,\!</math> then to compute the enlargement <math>\operatorname{E}q,</math> and finally to determine the difference <math>\operatorname{D}q = q + \operatorname{E}q,</math> so we let the variant order of terms reflect this sequence of considerations.
−Given these general considerations about the operators <math>\operatorname{E}</math> and <math>\operatorname{D},</math> let's return to particular cases, and carry out the first order analysis of the transformation <math>F(u, v) ~=~ ( ~\underline{((}~ u ~\underline{)(}~ v ~\underline{))}~ , ~\underline{((}~ u ~,~ v ~\underline{))}~ ).</math>
+Given these general considerations about the operators <math>\operatorname{E}</math> and <math>\operatorname{D},</math> let's return to particular cases, and carry out the first order analysis of the transformation <math>F(u, v) ~=~ ( ~\texttt{((u)(v))}~ , ~\texttt{((u,~v))}~ ).</math>
==Note 11==
==Note 11==
