X) ? 1, hence d X 1 (y) ? d Y (y) ? 1. Moreover d Y 1 (x) ? 2|Y | + 5 ? e(X) ? e(X) + 5 ? d X (x) + 5. Thus, e(X 1 , Y 1 ) = e(X, Y ) + d Y (y) ? d X 1 (y) + d X (x) ? d Y 1 (x) ? e(X, Y ,
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