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Examples of simplifying Boolean expression using De Morgan’s theorem

            Method 1: Breaking the long bar

                               A  BC

                                  Breaking long bar changes ORing (+) to ANDing


                             A BC

                                  Applying Complement law    A  A to  BC

                               ABC



            So,  A     reduces/simplifies to  ABC
            BC

            Method 2: Breaking the short bar first

                               A  BC


                                   Breaking the shortest bar changes ORing to ANDing

                            A  (B )     Applying the associative property to remove parenthesis



                                                                                     st
                            A  B  C  Breaking the long bar into two places between 1  and 2nd terms





                            A  B  C   and between 2nd and 3rd terms



                               ABC
















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