At the beginning of Section 12.6, it was noted that given the CBC MAC of a oneblock
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At the beginning of Section 12.6, it was noted that given the CBC MAC of a oneblock message \(X\), say \(T=\operatorname{MAC}(K, X)\), the adversary immediately knows the CBC MAC for the two-block message \(X \|(X \oplus T)\) since this is once again \(T\). Justify this statement.
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We use the definition from Section 126 For a oneblock mess...View the full answer
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