Consider the following Boolean expression. \(F=(X+Y+Z)(\overline X…
2021
Consider the following Boolean expression.
\(F=(X+Y+Z)(\overline X +Y)(\overline Y +Z)\)
Which of the following Boolean expressions is/are equivalent to \(\overline F\) (complement of \(F\))?
- A.
\((\overline X +\overline Y +\overline Z)(X+\overline Y)(Y+\overline Z)\) - B.
\(X\overline Y + \overline Z\) - C.
\((X+\overline Z)(\overline Y +\overline Z)\) - D.
\(X\overline Y +Y\overline Z + \overline X\; \overline Y \;\overline Z\)
Attempted by 194 students.
Show answer & explanation
Correct answer: B, C, D
Start by applying De Morgan's law to the product F = (X+Y+Z)(X̄+Y)(Ȳ+Z). The complement is a sum of complements:
(X+Y+Z)' = X̄ Ȳ Z̄
(X̄+Y)' = X Ȳ
(Ȳ+Z)' = Y Z̄
Therefore F' = X̄ Ȳ Z̄ + X Ȳ + Y Z̄
Show that X Ȳ + Z̄ is equivalent to this result:
X Ȳ + Z̄ = X Ȳ + (Ȳ Z̄ + Y Z̄) = X Ȳ + Y Z̄ + Ȳ Z̄
Ȳ Z̄ = X Ȳ Z̄ + X̄ Ȳ Z̄. The term X Ȳ Z̄ is absorbed by X Ȳ, leaving X̄ Ȳ Z̄.
Thus X Ȳ + Z̄ simplifies to X̄ Ȳ Z̄ + X Ȳ + Y Z̄, which is exactly F'.
Also note that (X+Z̄)(Ȳ+Z̄) expands and reduces to X Ȳ + Z̄, so it is equivalent to F'.
By contrast, (X̄ + Ȳ + Z̄)(X + Ȳ)(Y + Z̄) does not equal F'. For example, when X=1, Y=0, Z=1 this product evaluates to 0 while F' evaluates to 1, so the two expressions differ.
Final conclusion:
The following are equivalent to the complement of F: X Ȳ + Z̄
(X+Z̄)(Ȳ+Z̄) which reduces to X Ȳ + Z̄
X Ȳ + Y Z̄ + X̄ Ȳ Z̄ (the direct De Morgan form)
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