I am trying to do logic expression simplification using boolean algebra laws and axioms of logic, but I don't understand it at all. I attempt the question and this is what i have come up with.
the is the question giving : (¬a∨b)∧(a∨b)∧¬a
and this is what i got:
= (¬a∨b)∧(a∨b)∧¬a
= (¬a∨a)∧b)∧¬a - Distributive
=(¬a∨a)b¬a - Idempotent
=fvb -Complement or negation
=b identify
can someone please point me to the right direction and explain whats going on. i would be grateful thanks
You applied distributivity the wrong way.
Following the right distributive law (as JW Tanner points inn a comment) together with the commutative one, you get $$(a' \vee b) \wedge (a \vee b) \wedge a' = ((a' \wedge a) \vee b) \wedge a'.$$ (Here I'm using $x'$ for your notation $\neg x$).
Since $a' \wedge a = 0$ and $0 \vee b = b$, it all becomes $b \wedge a'$.
A different approach would be to use absorption: after using commutativity and associativity, you get $$(a' \vee b) \wedge (a \vee b) \wedge a' = ((a' \vee b) \wedge a') \wedge (a \vee b).$$ By absorption and distributivity, this gives $$a' \wedge (a \vee b) = (a' \wedge a) \vee (a' \wedge b) = 0 \vee (a' \wedge b) = a' \wedge b.$$