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Solved Examples · Example 15

Q.Convert Y=(A+B‾)(B+C)Y = (A + \overline{B})(B + C) into canonical POS form

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[!TLDR]

Fill in the missing variable in each OR term of Y=(A+B‾)(B+C)Y = (A + \overline{B})(B + C) so every sum term holds all of A, B and C, giving the four-maxterm canonical POS.

In a canonical (standard) POS expression each OR term must contain every input variable. Using XX‾=0X\overline{X} = 0, each sum term is ORed with XX‾X\overline{X} for its missing variable and then distributed by P+QR=(P+Q)(P+R)P + QR = (P + Q)(P + R).

Term (A+B‾)(A + \overline{B}): variable C is missing, so OR with CC‾C\overline{C}:

(A+B‾)+CC‾=(A+B‾+C)(A+B‾+C‾)(A + \overline{B}) + C\overline{C} = (A + \overline{B} + C)(A + \overline{B} + \overline{C})

Term (B+C)(B + C): variable A is missing, so OR with AA‾A\overline{A}:

(B+C)+AA‾=(A+B+C)(A‾+B+C)(B + C) + A\overline{A} = (A + B + C)(\overline{A} + B + C) …

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