Using set identities, prove that $\overline A \cup \overline B \cup (A \cap B \cap \overline C)= \overline A \cup \overline B \cup \overline C$

Question

I tried solving the problem but I get $U \cap U$ Which indeed $ \overline A \cup \overline B \cup \overline C$

Unfortunately I'm having a hard time proving it, any help?

Answer 1

You can use distributivity, note that:

$$\overline A \cup \overline B \cup (A \cap B \cap \overline C)= \overline A \cup \left[(\overline B \cup A)\cap(\overline B\cup B)\cap(\overline B\cup \overline C)\right].$$

Since obviously $\overline B\cup B=1$, then it equals

$$\overline A\cup\left[(\overline B \cup A)\cap(\overline B \cup \overline C)\right],$$

and using again distributivity:

$$\left[(\overline A \cup \overline B)\cup (\overline A \cup A)\right]\cap\left[(\overline A\cup \overline B)\cup(\overline A\cup\overline C)\right] =$$ $$= 1\cap (\overline A\cup \overline B \cup \overline C)=\boxed{\overline A\cup \overline B\cup \overline C}$$

Answer 2

Set $X=\overline{A}\cup\overline{B}$. Then from De Morgan's law you know that $$ \overline{X}=A\cap B $$ so your left-hand side becomes $$ X\cup(\overline{X}\cap\overline{C})=(X\cup\overline{X})\cap(X\cup\overline{C})=X\cup\overline{C} $$