boolean algebra Simplifying A'B'C'D+A'B'CD+A'BC'D'+A'BC'D+A'BCD'+AB'C'D'+AB'C'D+ABC'D'+ABCD

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boolean algebra Simplifying A'B'C'D+A'B'CD+A'BC'D'+A'BC'D+A'BCD'+AB'C'D'+AB'C'D+ABC'D'+ABCD

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-1 The relation R= (A,B,C,D,E) and functional dependencies F are given as follows: F= {A->BC, CD->E, B->D, E->A} E, BC and CD can be a candidate keys, but B cannot. Anyone could point me how this fact is calculated? I google it but couldn't understand more as what I known before. database database-design relational-database database-schema

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Using de morgan's law we can show that (A'B')' = A + B. NAND using OR & Inverter To implement NAND operation using OR gate, we first complement the inputs and then perform OR on the complemented inputs. We get A' + B'. A' + B' is equivalent to (AB)' which can be shown to be true by using de morgan's law. Procedure for Conversion

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Assume that the operators +, −, × are left associative and ^ is right associative. The order of precedence (form highest to lowest) is ^ , ×, +, −.The postfix expression corresponding to the infix expression a+b × d^e^f is

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