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DUE: 13 MAI 2026
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1. Let A, H, C be Sie ts and show t hat A U (B n C) = (A U B ) n (A U C)
Proof:
We will prove this by showing both inclusions:
1. A U(B nC)C(A UB)n(A. uC)
• Let X E A l J ( B n C)'
• This means that x E A or x E (B n C).
• lf x E A. then clearly z E (A U B ) and :x ,f (A LJ 0), so x E (A. u B ) n (A U C).
• If x E (B n C), then x E B at1d x E C, so x E ( A lJ B ) and x E: ( A LJ G), thus x E
(A u B ) n (A U C).
Hence, A U (B n C) C ( A LJ B ) n (A U C}.
2 . (A B ) n (A C) A B n C)
U lJ C U (
• (A lJ B ) n (A u C).
Let X E
• This means x E B) (A U:1: E a nd ( A U 0).
• If x E A. then clearly x E A LJ (B n C).
• lf x E B and .x E C, then x E (B n C). and thus x E A u (B r7 C).
Hence, (A lJ B ) n (A LJ 0) C A lJ (B n C).
Therefore, A u (B n C) =(A U B ) n {A u C).