1. 'Hence, base intruder, hence'
2. Without loss of generality, let b ? (B ? X), and suppose that b / ? (B ? Y ). If (B ? Y ) = ?, we have that X Y , so assume that (B ? Y ) = ?. Let c ? (B ? Y ). Now either b ? c or c ? b. If the former, then since Y is comprehensive, b ? (B ? Y ), which is false. Hence c ? b. As X is comprehensive;Suppose that is a benchmarking rule, with associated benchmarks B. First, suppose that B has the property that for all a, b ? B, either a ? b or b ? a