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Question. Name the law given and verify it using a truth table. X+ X’.Y=X+Y

My Answer. 
X   Y   X’  X’.Y    X+X’.Y  X+Y
0   0   1   0       0       0
0   1   1   1       1       1
1   0   0   0       1       1
1   1   0   0       1       1

Prove algebraically that X + X’Y = X + Y.
L.H.S. = X + X’Y
           = X.1 + X’Y        (X . 1 = X property of 0 and 1)
           = X(1 + Y) + X’Y   (1 + Y = 1 property of 0 and 1)
           = X + XY +  X’Y                                                             
           = X + Y(X + X’)
           = X + Y.1          (X + X’ =1 complementarity law)
           = X + Y            (Y . 1 = Y property of 0 and 1)
           = R.H.S.      Hence proved.

My teacher marked my answer wrong. And told me to find the correct answer. Friends tell me is it a complementary law or distributive law or Absorption law? If it is absorption kindly tell me how to prove RHS and LHS algebraically.

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    $\begingroup$ Your teacher may find your algebraic proof too complex (though it is correct, imho). For example, you use commutativity without need. Regarding the name of the law, I doubt very much it has one, but I would not underestimate the ability of people to create terminology just for the hell of it. $\endgroup$ – babou Aug 23 '15 at 13:30
  • $\begingroup$ We call it Absorption law, the prove is in any decent textbook. $\endgroup$ – Eugene Feb 20 '17 at 5:52
  • $\begingroup$ For me the absorption laws are $x \land (x \lor y) = x$ and $x \lor (x \land y) = x$. $\endgroup$ – Andrej Bauer Feb 20 '17 at 17:58
  • $\begingroup$ Did you talk to your teacher before seeking solace on the internet? $\endgroup$ – Andrej Bauer Feb 20 '17 at 17:58
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One way of looking at this is as a consequence of distributivity, where $P+QR\equiv (P+Q)(P+R)$. Then you'll have $$\begin{align} X+(X'Y) &\equiv (X+X')(X+Y)&\text{distributivity}\\ &\equiv T(X+Y)&\text{inverse}\\ &\equiv X+Y&\text{domination} \end{align}$$

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  • $\begingroup$ Thank you for your valuable answer. Is this equation any way is Absorption law ? If yes then how to prove RHS and LHS algebraically $\endgroup$ – user2241865 May 11 '14 at 7:02
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This is known as the third distributive law (see, e.g., this page). I think your proof is correct

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    $\begingroup$ Hmm. I've been doing this stuff for decades and I've never come across the term "third distributive law". $\endgroup$ – Rick Decker Jan 28 '16 at 1:51
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X+X'Y=X+Y

taking LHS X+X'Y=(X+X')(X+Y) OR distributes over AND 1.(X+Y) (X+X'=1) X+Y=RHS 1 is identity for AND

Hence Proved

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    $\begingroup$ I don't think this is the proof that the asker is looking for: it doesn't use the algebraic properties of the operators. Also, I find your proof very hard to read. What does "(X+Y)(X+X'=1)X+Y" mean? $\endgroup$ – David Richerby Mar 9 '16 at 16:22

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