Boolean Algebra Distribution

Question

I have some boolean algebra I preformed, but it does not appear to be right, and I am unable to justify to myself why.

$$F=(a+b+c)(a+b+d)$$ $$F'=(a'b'c')+(a'b'd')$$ $$F'=(a'b')(c'+d')$$ $$F=(a+b)(cd)$$

Yet if I let $a=c=d=0$ and $b=1$, then $$(a+b+c)(a+b+d)=(0+1+0)(0+1+0)=(1)(1)=1$$ and for the simplified equation, $$(a+b)(cd)=(0+1)(00)=(1)(0)=0$$ So it stands to reason that $$(a+b+c)(a+b+d) \ne (a+b)(cd)$$ but it seems like this should work. Would someone be able to explain where my mistake is?

Answer 1

When going from $$ F'=(a'b')(c'+d') $$ to $F$, it's rather $$ F=(a+b)\color{red}{+}cd. $$

Answer 2

The last step of your calculation is wrong:

$$\big((a'b')(c'+d')\big)'=(a'b')'+(c'+d')'=a+b+cd\;.$$