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To perform the chase test, first decompose all FD’s in ''F'' so each FD has a single attribute on the right hand side of the "arrow". ''F'' remains unchanged because all of its FD's already has a single attribute on the right hand side. ''F'' = {''A''→''B'', ''B''→''C'', ''CD''→''A''}.
When equating two symbols, if one of them is unsubscripted, make the other be the same so that the final tableau can have a row that is exactly the same as ''t'' = (''a'', ''b'', ''c'', ''d''). Also, if both have their own subscript, change either to be the other. However, to avoid confusion, all of the occurrences should be changed.
First, apply ''A''→''B'' to the tableau.
The first row is (''a'', ''b<sub>1</sub>'', ''c<sub>1</sub>'', ''d'') where ''a'' is unsubscripted and ''b<sub>1</sub>'' is subscripted with 1. Comparing the first row with the second one, change ''b<sub>2</sub>'' to ''b<sub>1</sub>''. Since the third row has ''a<sub>3</sub>'', ''b'' in the third row stays the same. The resulting tableau is: {| border="1" cellspacing="0" cellpadding="5" align="center"
! ''A'' !! ''B'' !! ''C'' !! ''D''
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