1
$\begingroup$

(A . B = A . (A̅ +B))

How to prove them? (Tried so many times)

5 Answers 5

1

A . (A̅ +B)

= A.A̅ + A.B

But A.A̅ = 0 ( Inverse or Complement Law)

= A.B

  • 0
    Mine pleasure :-) – 2017-02-01
  • 0
    Inverse Law and Complement law are same law but different names right? – 2017-02-01
  • 0
    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. – 2017-02-03
  • 0
    I got it. Complement law and inverse law are same law. See this link. http://www.electronics-tutorials.ws/boolean/bool_6.html – 2017-02-03
0

$$A\land (\lnot A \lor B)=(A\land\lnot A)\lor (A\land B)=0\lor (A\land B)=(A\land B)$$

  • 0
    can u explain the symbols dr. ∧, ∨, ¬, – 2017-02-01
  • 0
    These are the equivalent symbols in your notation: $$A\cdot B\sim A\land B,\quad A+B\sim A\lor B,\quad \overline{A}\sim \lnot A$$ – 2017-02-01
0

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.

0

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,

$$ A(\bar{A} + B) = A\cdot\bar{A} + AB = AB$$

  • 0
    distributive laws are : A (B + C) = AB + AC , A + (B . C) = (A + B) (A + C) right??. isn't this Inverse Law / Complement law ? – 2017-02-01
  • 0
    Just the first line was the application of distributive law. – 2017-02-01
  • 0
    i didn't get that dr. can u explain further. plz ?? – 2017-02-01
0

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.

  • 0
    thanks my friend. :) i think i messed up earlier. – 2017-02-03