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Question Bank (2 marks) · Q16

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

Karnataka PUCTextbookNumeric· 2mImportance★★★★★est
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[!TLDR]

Expanding each sum term to all three variables gives (A+C)(B+C‾)=∏M(0,1,2,5)(A+C)(B+\overline{C}) = \prod M(0,1,2,5).

A canonical (standard) POS expression is a product of maxterms — every sum term contains all the variables. The given expression Y=(A+C)(B+C‾)Y = (A + C)(B + \overline{C}) has sum terms missing one variable each, so each is expanded using the identity X+YZ=(X+Y)(X+Z)X + YZ = (X+Y)(X+Z), adding the missing variable as (variable)(variable‾)(\text{variable})(\overline{\text{variable}}).

Expand (A+C)(A + C) (missing B):

A+C=A+C+BB‾=(A+B+C)(A+B‾+C)A + C = A + C + B\overline{B} = (A + B + C)(A + \overline{B} + C) → maxterms M0,M2M_0, M_2.

Expand (B+C‾)(B + \overline{C}) (missing A): …

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