simplification of the function (boolean)

Question

$F(A, B, C, D) = (A'+B'+C')(A'+B')(A'+C')$

I am trying to simplify this and first tried doing something like: $(A'1+C)$ since $x'1$ is still $x$. I then did $1+C=1$, resulting in $A'1 = A'$, for that piece. I did the same for $(A'+B')$, however, is this allowed? If not, why? And what are the correct steps?

Answer 1

First use distribution $(P+Q)(P+R)(P+S)=P+(Q)(R)(S)$

Then use redundancy $(P+Q)\,Q=Q$

Then use something.

Answer 2

No, that's definitely not right. From $A'+C'$ you can indeed get $A'1+C'$, but that is not the same as $A'(1+C')$, which is what you are assuming. That would be like saying: $3\cdot 2 + 7=3 \cdot 9$