Question:
Simplify the Boolean expression:
\[
(x+y'+z')(x+y'+z)(x+y+z')
\]
Simplify the Boolean expression:
\[
(x+y'+z')(x+y'+z)(x+y+z')
\]
Show Hint
A powerful Boolean identity is:
\[
(A+B)(A+C)=A+BC
\]
It can reduce lengthy expressions in one step.
Updated On: Jun 11, 2026
- \(x+y'z'\)
- \(x+yz\)
- \(x'+y'z'\)
- \(xy'+z'\)
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
Use the Boolean identity:
\[
(A+B)(A+C)=A+BC
\]
to simplify products of sums.
Step 1: Simplify the first two factors.
Let \[ A=x+y' \] Then \[ (x+y'+z')(x+y'+z) \] becomes \[ (A+z')(A+z) \] Applying the identity, \[ (A+z')(A+z)=A \] Hence, \[ (x+y'+z')(x+y'+z)=x+y' \]
Step 2: Multiply with the remaining factor.
Now, \[ (x+y')(x+y+z') \] Again using \[ (A+B)(A+C)=A+BC \] with \[ A=x,\quad B=y',\quad C=y+z' \] we get \[ x+y'(y+z') \] \[ =x+y'y+y'z' \] \[ =x+0+y'z' \] \[ =x+y'z' \]
Step 3: Final simplified form.
\[ \boxed{x+y'z'} \]
Step 1: Simplify the first two factors.
Let \[ A=x+y' \] Then \[ (x+y'+z')(x+y'+z) \] becomes \[ (A+z')(A+z) \] Applying the identity, \[ (A+z')(A+z)=A \] Hence, \[ (x+y'+z')(x+y'+z)=x+y' \]
Step 2: Multiply with the remaining factor.
Now, \[ (x+y')(x+y+z') \] Again using \[ (A+B)(A+C)=A+BC \] with \[ A=x,\quad B=y',\quad C=y+z' \] we get \[ x+y'(y+z') \] \[ =x+y'y+y'z' \] \[ =x+0+y'z' \] \[ =x+y'z' \]
Step 3: Final simplified form.
\[ \boxed{x+y'z'} \]
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