Boolean algebra help with 3.12. and 3.14e. thank you in advance Boolean Algebra
ID: 3744947 • Letter: B
Question
Boolean algebra help with 3.12. and 3.14e.
thank you in advance
Explanation / Answer
3.12 A'CD'E+A'B'D'+ABCE+ABD=A'B'D'+ABD+BCD'E
A'CD'E(B+B')+A'B'D'+ABCE(D+D')+ABD=A'B'D'+ABD+BCD'E
A'CD'EB+A'CD'EB'+A'B'D'+ABCED'+ABCED+ABD=A'B'D'+ABD+BCD'E
A'B'D'(1+CE)+A'CD'EB+ABD(1+CE)+ABCED'=A'B'D'+ABD+BCD'E
A'B'D' +ABD +A'CD'EB+ABCED'=A'B'D'+ABD+BCD'E
A'B'D'+ABD+BCED'(A+A')=A'B'D'+ABD+BCD'E
A'B'D'+ABD+BCED'=A'B'D'+ABD+BCD'E
HENCE LHS =RHS
3.14 (e) WXY + WX'Y + WYZ + XYZ'
WY (X + X' + Z) + XYZ'
WY + XYZ'
Y (W + XZ')
Y (W + X) (W + Z')
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