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.

  • 3
    $\begingroup$ 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

One way of looking at this is as a consequence of distributivity, where $P+QR\equiv (P+Q)(P+R)$. Then you'll have $$\begin{align} X+(X'Y) &\equiv (X+X')(X+Y)&\text{distributivity}\\ &\equiv T(X+Y)&\text{inverse}\\ &\equiv X+Y&\text{domination} \end{align}$$

  • $\begingroup$ 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

This is known as the third distributive law (see, e.g., this page). I think your proof is correct

  • 2
    $\begingroup$ 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


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

  • 3
    $\begingroup$ 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

protected by David Richerby Jan 31 '18 at 15:56

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.