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Directory:Jon Awbrey/Papers/Dynamics And Logic (view source)
Revision as of 21:15, 12 June 2009
, 21:15, 12 June 2009→Note 3: markup
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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 f<p, q> over X = !P! x !Q!, the
(first order) "difference" of f is the proposition Df
over EX that is defined by the formula Df = Ef - f, or,
written out in full:
{| align="center" cellpadding="6" width="90%"
| align="center" |
<math>\begin{matrix}
\operatorname{D}f(p, q, \operatorname{d}p, \operatorname{d}q)
& = &
f(p + \operatorname{d}p,~ q + \operatorname{d}q) - f(p, q)
& = &
\texttt{(} f( \texttt{(} p, \operatorname{d}p \texttt{)},~ \texttt{(} q, \operatorname{d}q \texttt{)} ),~ f(p, q) \texttt{)}
\end{matrix}</math>
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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" |
<math>\begin{matrix}
\operatorname{D}f(p, q, \operatorname{d}p, \operatorname{d}q)
& = &
(p + \operatorname{d}p)(q + \operatorname{d}q) - pq
& = &
\texttt{((} p, \operatorname{d}p \texttt{)(} q, \operatorname{d}q \texttt{)}, pq \texttt{)}
\end{matrix}</math>
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| align="center" |
<pre>
o-------------------------------------------------o
o-------------------------------------------------o
| |
| |
| Df = ((p, dp)(q, dq), pq) |
| Df = ((p, dp)(q, dq), pq) |
o-------------------------------------------------o
o-------------------------------------------------o
</pre>
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<pre>
We did not yet go through the trouble to interpret this (first order)
We did not yet go through the trouble to interpret this (first order)
"difference of conjunction" fully, but were happy simply to evaluate
"difference of conjunction" fully, but were happy simply to evaluate