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Reduce the following Boolean expressions to the required number of literals.

(a) ABC+A'B'C+A'BC+ABC+A'B'C to five literals

(b) BC+AC'+AB+ BCD to four literals
                                                                  (c) [ (CD)' + A ]' +A+CD+ AB to three literals                                               
(d) (A + C + D)(A + C + D')(A + C + D)(A + B') to four literals
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2 Answers

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(a) A'B'C

(b) BC + AC' + AB + BCD

(c) (CD)'

(d) (A + C + D)(C + D')
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A literal is a variable or the complement of a variable.  

eg. AB+A’C → 4 literals – > A, A’, B, C,    AB + AC + D → 5 literals → A, B, A, C, D

 

a)     ABC + A’B’C + A’BC + ABC + A’B’C → ABC + A’B’C + A’BC → (take bc common from ABC and A’BC ) → 

                   (A+A’)BC + A’BC  → BC + A’BC  (No. of literals = 5 )

b)   BC + AC’ + AB + BCD →  BC (1 + D ) + AC’ + AB –> BC + AC’ + AB → use consensus property – > BC + AC’ (No of literals = 4)

​​​​​​​c)    [(CD)’ + A]’ + A + CD + AB → Use De Morgan’s in first term → ((CD)’)’ A’ + A +CD + AB → A’CD + A(1+B) + CD → 

                            A’CD + A + CD → CD(1+A’) + A → A + CD (No. of literals = 3)

​​​​​​​d)    (A+C+D)(A+C+D’)(A+C+D)(A+B’) → (A+C+D)(A+C+D’)(A+B’) → (A+C)(A+B’)  (No. of literals = 4 )

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