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Given frequent itemset l and subset s of l, prove that the confidence of the rule "s’ => (l - s’)" cannot be more than the confidence of "s => (l - s)"…

Given frequent itemset l and subset s of l, prove that the confidence of the rule "s’ => (l - s’)" cannot be more than the confidence of "s => (l - s)"…
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Given frequent itemset l and subset s of l, prove that the confidence of the rule "s’ => (l - s’)" cannot be more than the confidence of "s => (l - s)", where s’ is a subset of:_________

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User Funkymushroom
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Answer:

the answer is I

Explanation:

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User Simbu
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