Thank you!!! you dont know how thrilled i am!Acid_Snake wrote:lets use variable change so it becomes clear.3rdroyd wrote:(AB+C) * A'B = A'BAB + A'BC
NIce tut but...
I dont understand this!!!
I havn't fugure it out how to apply this property >>> A(B+C) = AB+AC on this one>>> (AB+C) * A'B = A'BAB + A'BC
So its (AB+C) to A'BAB ??? if true: Why A became a NOT A, (A') . Then why 2 B's and 2 A's >>> (A'BAB)???
im going to have a headache and want to understand this soooo badly, i think i will have nightmares tonight
We want to demonstrate that (AB+C) * A'B = A'BAB + A'BC
first thing I'm going to do is compare (AB+C) * A'B with (X+Y) * Z
therefore, we have:
X represents AB
Y represents C
Z represents A'B
since (X+Y) * Z is easier to see, we can use the property easier:
(X+Y) * Z = X*Z + Y*Z
Now we just change it back to the original values of X, Y and Z and we get:
X*Z + Y*Z -> AB*A'B + C*A'B
in other words: ABA'B + CA'B
so far so good, but this is not done yet.
If you look closely in the first AND we have ABA'B
if you do X AND X', the result is always 0
so:
ABA'B + CA'B = 0 + CA'B -> CA'B - > (we reorganize the variables) A'BC
the result result:
(AB+C) * A'B = A'BC
i understood when you exchange ABC to XYZ i can see it sooo clear now,
i also learned here, that to simplify fuctions you must! know all the properties very well otherwise you could get drowned in a glass of water
Thank you again!
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