MyWikiBiz, Author Your Legacy — Thursday November 07, 2024
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, 02:52, 27 May 2008
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| : <p><math>\operatorname{E}f(x, y, \operatorname{d}x, \operatorname{d}y) = xy + x\ \operatorname{d}y + y\ \operatorname{d}x + \operatorname{d}x\ \operatorname{d}y.</math> | | : <p><math>\operatorname{E}f(x, y, \operatorname{d}x, \operatorname{d}y) = xy + x\ \operatorname{d}y + y\ \operatorname{d}x + \operatorname{d}x\ \operatorname{d}y.</math> |
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− | To understand what this means in logical terms, for instance, as expressed in a boolean expansion or a ''disjunctive normal form'' (DNF), it is perhaps a little better to go back and analyze the expression the same way that we did for ''Df''. Thus, let us compute the value of the enlarged proposition ''Ef'' at each of the points in the universe of discourse ''U'' = ''X'' × ''Y''. | + | To understand what this means in logical terms, for instance, as expressed in a boolean expansion or a ''disjunctive normal form'' (DNF), it is perhaps a little better to go back and analyze the expression the same way that we did for <math>\operatorname{D}f.</math> Thus, let us compute the value of the enlarged proposition <math>\operatorname{E}f</math> at each of the points in the universe of discourse <math>U = X \times Y.</math> |
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| <pre> | | <pre> |
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| </pre> | | </pre> |
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− | Given the sort of data that arises from this form of analysis, we can now fold the disjoined ingredients back into a boolean expansion or a DNF that is equivalent to the proposition ''Ef''. | + | Given the sort of data that arises from this form of analysis, we can now fold the disjoined ingredients back into a boolean expansion or a DNF that is equivalent to the proposition <math>\operatorname{E}f.</math> |
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− | : ''Ef'' = ''xy Ef''<sub>''xy''</sub> + ''x''(''y'') ''Ef''<sub>''x''(''y'')</sub> + (''x'')''y Ef''<sub>(''x'')''y''</sub> + (''x'')(''y'') ''Ef''<sub>(''x'')(''y'')</sub> | + | : <p><math>\operatorname{E}f = xy\ \operatorname{E}f_{x y} + x (\!| y |\!)\ \operatorname{E}f_{x (\!| y |\!)} + (\!| x |\!) y\ \operatorname{E}f_{(\!| x |\!) y} + (\!| x |\!)(\!| y |\!)\ \operatorname{E}f_{(\!| x |\!)(\!| y |\!)}.</math></p> |
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| Here is a summary of the result, illustrated by means of a digraph picture, where the "no change" element (''dx'')(''dy'') is drawn as a loop at the point ''x y''. | | Here is a summary of the result, illustrated by means of a digraph picture, where the "no change" element (''dx'')(''dy'') is drawn as a loop at the point ''x y''. |