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→‎Note 3: markup
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<pre>
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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)
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"difference of conjunction" fully, but were happy simply to evaluate
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it with respect to a single location in the universe of discourse,
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namely, at the point picked out by the singular proposition pq,
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in as much as if to say at the place where p = 1 and q = 1.
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This evaluation is written in the form Df|pq or Df|<1, 1>,
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and we arrived at the locally applicable law that states
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that f = pq = p and q => Df|pq = ((dp)(dq)) = dp or dq.
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o-------------------------------------------------o
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{| align="center" cellpadding="6" width="90%"
|                                                |
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| align="center" |
|                                                 |
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<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>
|          o-----------o  o-----------o          |
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|-
|        /            \ /            \        |
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| align="center" |
|        /      p       o      q       \       |
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[[Image:Venn Diagram PQ Difference Conj At Conj.jpg|500px]]
|      /              /%\               \       |
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|-
|      /              /%%%\               \     |
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| align="center" |
|    o              o%%%%%o              o    |
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[[Image:Cactus Graph PQ Difference Conj At Conj.jpg|500px]]
|    |              |%%%%%|              |    |
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|}
|    |      dq (dp) |%%%%%|  dp (dq)      |    |
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|    |  o<----------|--o--|---------->o  |    |
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|    |              |%%|%%|              |    |
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|    |              |%%|%%|              |    |
  −
|    o              o%%|%%o              o    |
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|      \               \%|%/              /      |
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|      \               \|/              /      |
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|        \               |              /       |
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|         \            /|\            /        |
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|          o-----------o | o-----------o          |
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|                       |                        |
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|                      dp|dq                      |
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|                        |                        |
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|                        v                        |
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|                       o                        |
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|                                                 |
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o-------------------------------------------------o
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|                                                 |
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|                     dp  dq                    |
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|                     o  o                      |
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|                      \ /                      |
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|                        o                        |
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|                        |                        |
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|                        @                        |
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|                                                |
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o-------------------------------------------------o
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| Df|pq =          ((dp) (dq))                  |
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o-------------------------------------------------o
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The picture illustrates the analysis of the inclusive
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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:
disjunction ((dp)(dq)) into the exclusive disjunction:
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dp(dq) + (dp)dq + dp dq, a differential proposition that
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{| align="center" cellpadding="6" width="90%"
may be interpreted to say "change p or change q or both".
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| align="center" |
And this can be recognized as just what you need to do if
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<math>\begin{matrix}
you happen to find yourself in the center cell and require
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\operatorname{d}p ~\texttt{(} \operatorname{d}q \texttt{)}
a complete and detailed description of ways to escape it.
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& + &
</pre>
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\texttt{(} \operatorname{d}p \texttt{)}~ \operatorname{d}q
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& + &
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\operatorname{d}p ~\operatorname{d}q
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\end{matrix}</math>
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|}
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The differential proposition that results may be interpreted to say "change <math>p\!</math> or change <math>q\!</math> or both". And this can be recognized as just what you need to do if you happen to find yourself in the center cell and require a complete and detailed description of ways to escape it.
    
==Note 4==
 
==Note 4==
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