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3.11ABBA030409

# 3.11ABBA030409 - 1 3.11 ABBA Recall the reduced AB table...

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1 3.11. ABBA. Recall the reduced AB table from section 3.5. REDUCED AB 1. A . fA B . gA. INF. AL. ALF. FIN. NON. 2. A . fA B . gB. INF. AL. ALF. FIN. NON. 3. A . fA B . gC. INF. AL. ALF. FIN. NON. 4. C . fA B . gA. INF. AL. ALF. FIN. NON. 5. C . fA B . gB. INF. AL. ALF. FIN. NON. 6. C . fA B . gC. INF. AL. ALF. FIN. NON. Recall the reduced BA table from section 3.6. REDUCED BA 1’. B . fB A . gB. INF. AL. ALF. FIN. NON. 2’. B . fB A . gA. INF. AL. ALF. FIN. NON. 3’. B . fB A . gC. INF. AL. ALF. FIN. NON. 4’. C . fB A . gB. INF. AL. ALF. FIN. NON. 5’. C . fB A . gA. INF. AL. ALF. FIN. NON. 6’. C . fB A . gC. INF. AL. ALF. FIN. NON. This results in 36 ordered pairs. We can take advantage of symmetry through interchanging A with B as follows. Clearly (i,j’) and (j,i’) are equivalent, since interchanging A and B takes us from p to p’ and back. So we can require that i j. Thus we have the following 21 ordered pairs to consider. We need to determine the status of all attributes INF, Al, ALF, FIN, NON, for each pair.

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