Convert the following Boolean expression into its canonical POS form:F

1.	Convert the following Boolean expression into its canonical POS form:F(A, B, C) = (B + C’).(A’ + B)
Answer» X + YZ = (X + Y)(X + Z) Now A.A’ = 0, Similarly, C.C’ = 0 Therefore, (B + C’) = (B + C’ + A.A’) = (B + C + A) (B + C + A’) [Treating B + C’ as a single variable] and(A’ + B) = (A’+B + C.C’) = (A’ + B + C)(A’ + B + C’) [Treating A’ + B as a single variable] F(A, B, C) = (B + C’).(A’ + B) = (B + C’ + A)(B + C’ + A’)(A’ + B + C)(A’ + B + C’)

Convert the following Boolean expression into its canonical POS form:F(A, B, C) = (B + C’).(A’ + B)

Answer»

X + YZ = (X + Y)(X + Z)

Now A.A’ = 0, Similarly, C.C’ = 0

Therefore, (B + C’) = (B + C’ + A.A’) = (B + C + A) (B + C + A’)

[Treating B + C’ as a single variable]

and(A’ + B) = (A’+B + C.C’) = (A’ + B + C)(A’ + B + C’)

[Treating A’ + B as a single variable]

F(A, B, C) = (B + C’).(A’ + B)

= (B + C’ + A)(B + C’ + A’)(A’ + B + C)(A’ + B + C’)