Prove that $A' ∪ B' ∪ (A ∩ B ∩ C')$ is equivalent to $A' ∪ B' ∪ C'$ by algebraic laws

Question

I know $A' ∪ B' ∪ (A ∩ B ∩ C')$ is equivalent to $A' ∪ B' ∪ C'$ by drawing Venn diagram.

But I can't successfully prove by algebraic laws.

Here is what I've tried:

$$A' ∪ B' ∪ (A ∩ B ∩ C')$$ $$= A' ∪ [(B' ∪ A) ∩ (B' ∪ B) ∩ (B' ∪ C')] \quad\text{distributive laws}$$ $$= A' ∪ [(B' ∪ A) ∩ T ∩ (B' ∪ C')] \quad\text{negation laws} $$ $$= A' ∪ [(B' ∪ A) ∩ (B' ∪ C')] \quad\text{identity laws} $$

And then, I can't proceed to $A' ∪ B' ∪ C'$.

Answer 1

I think you only need to apply the distributive law once. $$\begin{array}{rl} A'\cup B'\cup\left(A\cap B \cap C'\right) = & (A'\cup B')\cup\left[(A\cap B) \cap C'\right] \mbox{ associativity }\\ = & [(A'\cup B')\cup(A\cap B)] \cap (A'\cup B'\cup C')\mbox{ distributivity }\\ = & A'\cup B'\cup C' \mbox{ negation of intersection} \end{array}$$