(a∧–b) ∨ (–b∧c) ∨ (–a∧c) = (a∧–b) ∨ (–a∧c)
How can I prove this,
I tried to use the distribution law for the first/last two terms, but it doesn't seem to work. Can anyone give me some hints?
Thanks a lot
If you will remember the algebra of sets is Boolean isomorphic to sentence logic. So, this is equivalent to showing that
$\displaystyle \left(A\cap B'\right)\cup\left(B'\cap C\right)\cap\left(A'\cap C\right)=\left(A\cap B'\right)\cup\left(A'\cap C\right)$
The left side is $\displaystyle \left(A\cap B'\right)\cup\left(B'\cap C\right)=\left(A\cap\left(B'\cap C\right)\right)\cup\left(B'\cap\left(B'\cap C\right)\right)=\left(A\cap B'\cap C\right)\cup\left(B'\cap C\right)$$\displaystyle =B'\cap C$
So the left side is $\displaystyle \left(B'\cap C\right)\cap\left(A'\cap C\right)=B'\cap A'\cap C$ and the right side is $\displaystyle \left(A\cap B'\right)\cup\left(A'\cap C\right)=\left(A'\cup\left(A\cap B'\right)\right)\cap\left(\left(C\cup\left(A\cap B'\right)\right)\right)$ which equals $\displaystyle \left(\left(A'\cup A\right)\cap\left(A'\cap B'\right)\right)\cap\left(\left(C\cup A\right)\cap\left(C\cap B'\right)\right)=\left(A'\cap B'\right)\cap\left(C\cap A\cap B'\right)$ which gives...bleh.
OK, here is what I did. First I checked using truth tables that the two sides are equal. I have not constructed the whole truth table but looked at when one side is false, which means that all disjuncts are false.(a /\ -b) \/ (-b /\ c) \/ (-a /\ c) = (a /\ -b) \/ (-a /\ c)
Next, it's easy to see that the left-hand side is the right-hand side plus one disjunct: (-b /\ c). So we should show that adding this disjunct does not change anything. If we show that (a /\ -b) \/ (-a /\ c) = (-b /\ c) \/ X for some X, then
(a /\ -b) \/ (-b /\ c) \/ (-a /\ c) = (-b /\ c) \/ X \/ (-b /\ c) = (-b /\ c) \/ X = (a /\ -b) \/ (-a /\ c)
To show that (a /\ -b) \/ (-a /\ c) = (-b /\ c) \/ X, we can use the fact that the representation of a function as a DNF where each disjunct has all prop. letters is unique. So one has to add c to the first disjunct on the left and b to the second one, as well as a to the right-hand side. This can be done by adding c \/ -c, b \/ -b and a \/ -a and using the distributive law.