so I have been trying to learn simplifying boolean Algebra, lets look at this term : $$ (a*b)+(a*\neg b)+(\neg a * \neg b)$$
I have several questions, I know how the right solution looks like but why couldn't I collect $a$ and $\neg a$?
so :
1. $a* \neg a *(b+\neg b+\neg b)$
2. $0*(1)$
is the way I simplifyied it right, is it just non-sense doing it like this or what is I wrong with this?
For the derivation, notice that there are 4 possible states of $a$ and $b$:
$$a = 0, b = 0 \\ a=0, b=1 \\ a=1,b=0 \\ a=1, b=1$$
Now, the expression $(a*b)$ is true for the fourth state, the expression $(a*\neg b)$ is true for the third state and the expression $(\neg a * \neg b)$ is true for the first state. So your expression is equivalent to: $$(a*b)+(a*\neg b)+(\neg a * \neg b) \equiv \neg(\neg a * b) $$ ("Just not the second option!"). By De-Morgan's rule: $$\neg(\neg a * b) \equiv a + \neg b$$