(A . B = A . (A̅ +B))
How to prove them? (Tried so many times)
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A . (A̅ +B)
= A.A̅ + A.B
But A.A̅ = 0 ( Inverse or Complement Law)
= A.B
Kanwaljit Singh
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Mine pleasure :-) – Kanwaljit Singh Feb 01 '17 at 09:58
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Inverse Law and Complement law are same law but different names right? – Bachi Nirosh Feb 01 '17 at 10:07
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dr. i saw on two difference web sites but same formula. they say each both names that i mentioned (Inverse Law and Complement law ). can u plz teach me differences between those 2 formulas. – Bachi Nirosh Feb 03 '17 at 03:56
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I got it. Complement law and inverse law are same law. See this link. http://www.electronics-tutorials.ws/boolean/bool_6.html – Kanwaljit Singh Feb 03 '17 at 10:40
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$$A\land (\lnot A \lor B)=(A\land\lnot A)\lor (A\land B)=0\lor (A\land B)=(A\land B)$$
b00n heT
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These are the equivalent symbols in your notation: $$A\cdot B\sim A\land B,\quad A+B\sim A\lor B,\quad \overline{A}\sim \lnot A$$ – b00n heT Feb 01 '17 at 10:03
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Hint: Use the fact that the product of a variable and its complement is $0$. Thus, $AA' =0$. Can you take it from here? Hope it helps.
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Use the distributive law and you get
$$ A\cdot\bar{A} + A\cdot B. $$
$\text{True AND False}$ results in $\text{False}$. Which give $A\cdot \bar{A} = 0$. Again, $\text{False OR X}$ always results in $X$, hence we get $0 + AB = AB$. Thus,
$$ A(\bar{A} + B) = A\cdot\bar{A} + AB = AB$$
Nilabro Saha
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distributive laws are : A (B + C) = AB + AC ,
A + (B . C) = (A + B) (A + C) right??. isn't this Inverse Law / Complement law ? – Bachi Nirosh Feb 01 '17 at 10:10 -
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A•(Ā + B) = A•Ā + A•B using the distributive property.
EDIT: This is in response to the OP's request to explain further. Wanted to add this as a comment, but encountered an error.
Nilabro Saha
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