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Revision as of 02:34, 21 June 2009
, 02:34, 21 June 2009→Note 3: convert graphics
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−<pre>+o-------------------------------------------------o+| |
−| p dp q dq |
−| o---o o---o |
−| \ | | / |
−| \ | | / |
−| \| |/ p q |
−| o=o-----------o |
−| \ / |
−| \ / |
−| \ / |
−| \ / |
−| \ / |
−| \ / |
−| @ |
−| |
−o-------------------------------------------------o
−o-------------------------------------------------o
−</pre>
following formula:
following formula:
−{| align="center" cellpadding="6" width="90%"
+{| align="center" cellspacing="10" width="90%"
| align="center" |
| align="center" |
<math>\begin{matrix}
<math>\begin{matrix}
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In the example <math>f(p, q) = pq,\!</math> the enlargement <math>\operatorname{E}f</math> is computed as follows:
In the example <math>f(p, q) = pq,\!</math> the enlargement <math>\operatorname{E}f</math> is computed as follows:
−{| align="center" cellpadding="6" width="90%"
+{| align="center" cellspacing="10" width="90%"
| align="center" |
| align="center" |
<math>\begin{matrix}
<math>\begin{matrix}
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Given the proposition <math>f(p, q)\!</math> over <math>X = P \times Q,</math> the ''(first order) difference'' of <math>f\!</math> is the proposition <math>\operatorname{D}f</math> over <math>\operatorname{E}X</math> that is defined by the formula <math>\operatorname{D}f = \operatorname{E}f - f,</math> or, written out in full:
Given the proposition <math>f(p, q)\!</math> over <math>X = P \times Q,</math> the ''(first order) difference'' of <math>f\!</math> is the proposition <math>\operatorname{D}f</math> over <math>\operatorname{E}X</math> that is defined by the formula <math>\operatorname{D}f = \operatorname{E}f - f,</math> or, written out in full:
−{| align="center" cellpadding="6" width="90%"
+{| align="center" cellspacing="10" width="90%"
| align="center" |
| align="center" |
<math>\begin{matrix}
<math>\begin{matrix}
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In the example <math>f(p, q) = pq,\!</math> the difference <math>\operatorname{D}f</math> is computed as follows:
In the example <math>f(p, q) = pq,\!</math> the difference <math>\operatorname{D}f</math> is computed as follows:
−{| align="center" cellpadding="6" width="90%"
+{| align="center" cellspacing="10" width="90%"
| align="center" |
| align="center" |
<math>\begin{matrix}
<math>\begin{matrix}
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\texttt{((} p, \operatorname{d}p \texttt{)(} q, \operatorname{d}q \texttt{)}, pq \texttt{)}
\texttt{((} p, \operatorname{d}p \texttt{)(} q, \operatorname{d}q \texttt{)}, pq \texttt{)}
\end{matrix}</math>
\end{matrix}</math>
−|-
+|}
−| align="center" |
+{| align="center" cellspacing="10"
−| [[Image:Cactus Graph Df = ((P,dP)(Q,dQ),PQ).jpg|500px]]
−| Df = ((p, dp)(q, dq), pq) |
−|}
|}
We did not yet go through the trouble to interpret this (first order) ''difference of conjunction'' fully, but were happy simply to evaluate it with respect to a single location in the universe of discourse, namely, at the point picked out by the singular proposition <math>pq,\!</math> that is, at the place where <math>p = 1\!</math> and <math>q = 1.\!</math> This evaluation is written in the form <math>\operatorname{D}f|_{pq}</math> or <math>\operatorname{D}f|_{(1, 1)},</math> and we arrived at the locally applicable law that is stated and illustrated as follows:
We did not yet go through the trouble to interpret this (first order) ''difference of conjunction'' fully, but were happy simply to evaluate it with respect to a single location in the universe of discourse, namely, at the point picked out by the singular proposition <math>pq,\!</math> that is, at the place where <math>p = 1\!</math> and <math>q = 1.\!</math> This evaluation is written in the form <math>\operatorname{D}f|_{pq}</math> or <math>\operatorname{D}f|_{(1, 1)},</math> and we arrived at the locally applicable law that is stated and illustrated as follows:
−{| align="center" cellpadding="6" width="90%"
+{| align="center" cellspacing="10" width="90%"
| align="center" |
| align="center" |
<math>f(p, q) ~=~ pq ~=~ p ~\operatorname{and}~ q \quad \Rightarrow \quad \operatorname{D}f|_{pq} ~=~ \texttt{((} \operatorname{dp} \texttt{)(} \operatorname{d}q \texttt{))} ~=~ \operatorname{d}p ~\operatorname{or}~ \operatorname{d}q</math>
<math>f(p, q) ~=~ pq ~=~ p ~\operatorname{and}~ q \quad \Rightarrow \quad \operatorname{D}f|_{pq} ~=~ \texttt{((} \operatorname{dp} \texttt{)(} \operatorname{d}q \texttt{))} ~=~ \operatorname{d}p ~\operatorname{or}~ \operatorname{d}q</math>
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The picture shows the analysis of the inclusive disjunction <math>\texttt{((} \operatorname{d}p \texttt{)(} \operatorname{d}q \texttt{))}</math> into the following exclusive disjunction:
The picture shows the analysis of the inclusive disjunction <math>\texttt{((} \operatorname{d}p \texttt{)(} \operatorname{d}q \texttt{))}</math> into the following exclusive disjunction:
−{| align="center" cellpadding="6" width="90%"
+{| align="center" cellspacing="10" width="90%"
| align="center" |
| align="center" |
<math>\begin{matrix}
<math>\begin{matrix}
