Tazmin Horton

2021-02-05

Which one is a simplified expression for $A\left({A}^{\prime }+{B}^{\prime }\right)\left(B+C\right)\left(B+{C}^{\prime }+D\right)$

Usamah Prosser

Step 1
Multiply A with the terms in the first bracket:
$=\left[A\left({A}^{\prime }+{B}^{\prime }\right)\right]\left(B+C\right)\left(B+{C}^{\prime }+D\right)$
$=\left[{\mathrm{\forall }}^{\prime }+A{B}^{\prime }\right]\left(B+C\right)\left(B+{C}^{\prime }+D\right)$
Since $P{P}^{\prime }=0$
$=\left[0+A{B}^{\prime }\right]\left(B+C\right)\left(B+{C}^{\prime }+D\right)$
$=A{B}^{\prime }\left(B+C\right)\left(B+{C}^{\prime }+D\right)$
Step 2
Multiply AB' with the terms in the first bracket of the obtained expression:
$=\left[A{B}^{\prime }\left(B+C\right)\right]\left(B+{C}^{\prime }+D\right)$
$=\left[AB{B}^{\prime }+A{B}^{\prime }C\right]\left(B+{C}^{\prime }+D\right)$
$=\left[A\left(B{B}^{\prime }\right)+A{B}^{\prime }C\right]\left(B+{C}^{\prime }+D\right)$
Since $P{P}^{\prime }=0$
$=\left[0+A{B}^{\prime }C\right]\left(B+{C}^{\prime }+D\right)$
$=A{B}^{\prime }C\left(B+{C}^{\prime }+D\right)$
Step 3
Multiply AB'C with all the terms in the bracket of the obtained expression:
$=A{B}^{\prime }C\left(B+{C}^{\prime }+D\right)$
$=AB{B}^{\prime }C+A{B}^{\prime }C{C}^{\prime }+A{B}^{\prime }CD$
$=A\left(B{B}^{\prime }\right)C+A{B}^{\prime }\left(C{C}^{\prime }\right)+A{B}^{\prime }CD$
Since $P{P}^{\prime }=0$
$=0+0+A{B}^{\prime }CD$
$=A{B}^{\prime }CD$
Thus the simplified expression is AB'CD.

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