, We show that if there exists a table S where I(p1, S) ? I(p2, S), then p1 p2. For showing p1 p2, we define a constrained table T = (D, R) such that I(p2, R) ? I(p2, D) and I(p1, R) ? I(p1, D)

, Now we have to show that I(p1, R) ? I(p1, D)

S. I(p2 and S. , we conclude I(p1, R) ? I(p2, R) and it is sufficient to show that I(p1, D) ? I(p2, R): I(p1, D) = I(p1, I(p2, R)) ? I

, On the other hand, by proposition 6, we know that p1 p2 implies ?S : I(p1, S) ? I(p2, S) (contradiction). We now show that if p1 is a specialization of p2, then p1 p2, vol.1

, Proof of Proposition 2. By contradiction using the notion of cover and subsumption. Suppose that there exist two minimal strict covers P * (T )1 and P * (T )2. Then there exists a pattern p1 ? P * (T )1 ? P * (T )2 and a

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