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A'B'D+ A'BD + BCD + ACD Boolean alegbra?

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  • A'B'D+ A'BD + BCD + ACD Boolean alegbra?


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Answer #1 | 28/12 2013 23:08
You can factor the sentence to D(A'B'+A'B+BC+AC) or D[A'(B'+B)+C(B+A)]=
Answer #2 | 29/12 2013 01:20
D(A'B'+A'B+BC+AC) D(A'(B'+B) + C(B+A)) Since (B'+B) = true always D(A' + C(B+A))
Answer #3 | 29/12 2013 08:38
A'B'D+ A'BD + BCD + ACD A'D (B' + B) + BCD + ACD <<== T10 (Theorem 10 in attached source): A + A' = 1 A'D + BCD + ACD D(A' + BC + AC) <<== Extract common term D(A' (1 + C) + BC + AC) <<== T8: A + 1 = 1 D(A' + A'C + BC + AC) D(A' + BC + C (A + A')) <<== T10 D(A' + BC + C) D(A' + C (1 + B)) <<== T8 D(A' + C) <<== Either A'D + CD <<== or

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