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==Note 13==
 
==Note 13==
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I think that it ought to be clear at this point that we need a more systematic symbolic method for computing the differentials of logical transformations, using the term ''differential'' in a loose way at present for all sorts of finite differences and derivatives, leaving it to another discussion to sharpen up its more exact technical senses.
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It ought to be clear at this point that we need a more systematic symbolic method for computing the differentials of logical transformations, using the term ''differential'' in a loose way at present for all sorts of finite differences and derivatives, leaving it to another discussion to sharpen up its more exact technical senses.
    
For convenience of reference, let's recast our current example in the following form:
 
For convenience of reference, let's recast our current example in the following form:
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| <math>F ~=~ (f, g) ~=~ ( ~\underline{((}~ u ~\underline{)(}~ v ~\underline{))}~, ~\underline{((}~ u ~,~ v ~\underline{))}~ ).</math>
 
| <math>F ~=~ (f, g) ~=~ ( ~\underline{((}~ u ~\underline{)(}~ v ~\underline{))}~, ~\underline{((}~ u ~,~ v ~\underline{))}~ ).</math>
 
|}
 
|}
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In their application to this logical transformation the operators <math>\operatorname{E}</math> and <math>\operatorname{D}</math> respectively produce the ''enlarged map'' <math>\operatorname{E}F = (\operatorname{E}f, \operatorname{E}g)</math> and the ''difference map'' <math>\operatorname{D}F = (\operatorname{D}f, \operatorname{D}g),</math> whose components can be given as follows, if the reader, in the absence of a special format for logical parentheses, can forgive syntactically bilingual phrases:
    
<pre>
 
<pre>
In their application to this logical transformation the operators
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E and D respectively produce the "enlarged map" EF = <Ef, Eg> and
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the "difference map" DF = <Df, Dg>, whose components can be given
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as follows, if the reader, in the absence of a special format for
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logical parentheses, can forgive syntactically 2-lingual phrases:
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Ef  =  ((u + du)(v + dv))
 
Ef  =  ((u + du)(v + dv))
  
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