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Can anyone help me with the following Boolean Algebra expression?

$Y\hspace{1mm}=\hspace{1mm}\overline{A}B\overline{C}D\hspace{1mm}+\hspace{1mm}\overline{A}BC\overline{D}\hspace{1mm}+\hspace{1mm}\overline{A}BCD\hspace{1mm}+\hspace{1mm}AB\overline{C}D\hspace{1mm}+\hspace{1mm}ABC\overline{D}\hspace{1mm}+\hspace{1mm}ABCD$

I have been trying to solve it but I can't seem to the answer which is $BC + BD$. I got this answer using K-Map and I just want to verify it using the Algebraic method. This is what I have been doing

My Method

Am I doing anything wrong?
Thank You

Utkarsh
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1 Answers1

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First note that the six products fall into three pairs that differ in only the $A$ literal: $$(B\overline CD)(A+\overline A)+(BC\overline D)(A+\overline A)+(BCD)(A+\overline A)$$ Since $A+\overline A=1$, this reduces to $$B\overline CD+BC\overline D+BCD=B(CD+\overline CD+C\overline D)$$ Now $$CD+\overline CD+C\overline D=C(D+\overline D)+\overline CD=C+\overline CD$$ $$=\overline{\overline C(\overline D+C)}=\overline{\overline C\cdot\overline D}=C+D$$ Eventually we reach the desired result of $B(C+D)$.

Parcly Taxel
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