![\"Boolean](\"https:\/\/bcisnotes.com\/secondsemester\/wp-content\/uploads\/2019\/07\/a-4.png\")
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A Boolean function may be transformed from an algebraic expression into a logic diagram composed of AND, OR, and NOT gates. The implementation of the four functions introduced in the previous discussion is shown below:<\/p>\n
![\"Boolean](\"https:\/\/bcisnotes.com\/secondsemester\/wp-content\/uploads\/2019\/07\/b-3-1024x808.png\")
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Algebraic manipulation and simplification of Boolean function <\/strong><\/p>\n <\/p>\n \u00a0Simplify the following Boolean functions to a minimum number of literals.<\/strong><\/p>\n <\/p>\n <\/p>\n The Complement of a function <\/strong><\/p>\n The complement of a function F <\/em>is F’ <\/em>and is obtained from an interchange of 0’s for 1’s and 1’s for 0’s in the value of F. <\/em>The complement of a function may be derived algebraically through DeMorgan’s theorem. DeMorgan’s theorems can be extended to three or more variables. The three-variable form of the first DeMorgan’s theorem is derived below.<\/p>\n (A + B + C)’ = (A + X)’ let B + C = X<\/p>\n = A’X’ (DeMorgan)<\/p>\n = A’\u00b7 (B + C)’ substitute B + C = X<\/p>\n = A’.(B ‘C’) (DeMorgan)<\/p>\n = A’B’C’ (associative)<\/p>\n DeMorgan’s theorems for any number of variables resemble in form the two-variable case and can be derived by successive substitutions similar to the method used in the above derivation. These theorems can be generalized as follows:<\/p>\n (A + B + C + D + … + F)’ = A’B’C’D’\u2026 F’<\/p>\n (ABCD … F)’ = A’ + B’ + C’ + D’ + … + F’<\/p>\n Algebraic manipulation and simplification of a Boolean function <\/strong><\/p>\n <\/p>\n Simplify the following Boolean functions to a minimum number of literals.<\/p>\n <\/p>\n The complement of a function F <\/em>is F’ <\/em>and is obtained from an interchange of 0’s for 1’s and 1’s for 0’s in the value of F. <\/em>The complement of a function may be derived algebraically through DeMorgan’s theorem. DeMorgan’s theorems can be extended to three or more variables. The three-variable form of the first DeMorgan’s theorem is derived below.<\/p>\n (A + B + C)’ = (A + X)’ let B + C = X<\/p>\n = A’X’ (DeMorgan)<\/p>\n = A’\u00b7 (B + C)’ substitute B + C = X<\/p>\n = A’.(B’C’) (DeMorgan)<\/p>\n = A’B’C’ (associative)<\/p>\n DeMorgan’s theorems for any number of variables resemble in form the two variable case and can be derived by successive substitutions similar to the method used in the above derivation. These theorems can be generalized as follows:<\/p>\n (A + B + C + D + … + F)’ = A’B’C’D’\u2026 F’<\/p>\n (ABCD … F)’ = A’ + B’ + C’ + D’ + … + F’<\/p>\n You may also like the basic theorems & properties of boolean algebra.<\/a><\/p>\n\n
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\n= xy + x’z + xyz + x’yz
\n= xy(1 + z) + x’z(1 + y)
\n= xy + x’z<\/li>\n<\/ol>\n\n
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\n= xy + x’z + xyz + x’yz
\n= xy(1 + z) + x’z(1 + y)
\n= xy + x’z<\/li>\n<\/ol>\n\n
A Complement of a function<\/strong><\/h2>\n