Prove that (a+bc) + (a'(b'+c')) = 1

Question

Im trying to prove the boolean expression (a+bc) is the complement of (a'(b'+c')). To do that I need to prove that (a+bc) + (a'(b'+c')) = 1, however I only get as far as:

(a + bc + a')(a + bc + b' + c')
(a + a' + bc)(a + bc + b' + c')
(1 + bc)(a + bc + b' + c')
1(a + bc + b' + c')

What property can I use to reduce the RHS?

Answer 1

You're doing very well! Now, recall that $$b'+c'=(bc)',$$ so that we get $$a+bc+(bc)'.$$ Can you take it from there?

Answer 2

$$a \lor bc \lor \overline a(\overline b \lor \overline c)=1$$ I use this rule: $a=a \lor a \overline b=a \lor a \overline c$. $$[a \lor bc \lor \overline a \overline b \lor \overline a \cdot \overline c] \lor a \overline b \lor a \overline c =1 $$ You should remember that $ x \lor \overline x = 1$. $$ a \lor \overline b \lor \overline c \lor bc =1$$ $$ a \lor \overline b \lor \overline c \lor bc \lor \overline b c =1$$ $$ a \lor \overline b \lor \overline c \lor c =1$$ $$ a \lor \overline b \lor 1 =1$$ So, this function is always true.