0
$\begingroup$

Hi guys can you help me with the following task:

  1. Using the laws of Boolean algebra, show the validity of the following expression:

$ (x \land y) \lor (x \land \overline y) = x $

I have done:

$ (x \land y) \lor (x \land \overline y) = x \land (\overline y \lor y) = x $

  1. Which new law results from the application of the duality principle of Boolean algebra? algebra? Check the validity of the new law with the calculation rules of Boolean algebra (as above)

I don't understand which law follows from that.

$\endgroup$
2
  • 1
    $\begingroup$ Apply the rules of Duality principle in boolean algebra: in the specific case, swap $\lor$ and $\land$ and check if the result is still valid. $\endgroup$ Commented Nov 15, 2021 at 15:40
  • $\begingroup$ The dual of $ (x \land y) \lor (x \land \overline y) = x $ is $ (x \lor y) \land (x \lor \overline y) = x $.. $\endgroup$ Commented Nov 16, 2021 at 14:39

1 Answer 1

Reset to default
1
$\begingroup$

Note that almost all laws in boolean algebra come in pairs:

For example:

Commutation

$x \land y = y \land x$

$x \lor y = y \lor x$

Distribution

$x \land (y \lor z) = (z \land y) \lor (x \land z)$

$x \lor (y \land z) = (z \lor y) \land (x \lor z)$

DeMorgan

$\neg(x \land y) = \neg x \lor \neg y$

$\neg(x \lor y) = \neg x \land \neg y$

Etc.

You'll note that when we have one of the two laws, we can always get the other one by systematically switching $\land$ and $\lor$. This is not a coincidence: The Duality Theorem is a general theorem that shows that this works for any equivalence principle. So ... take your just-found equivalence, and find its partner.

Moreover, once you have that dual law, you'll find that its derivation uses the same (but dual) laws as the derivation you did for 1 (good job on number 1 by the way)

$\endgroup$
1
  • 1
    $\begingroup$ $0$ and $1$ (if they appear) also need to be swapped. $\endgroup$ Commented Nov 16, 2021 at 14:42

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.