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Is the function $(p \wedge q) \vee r$ equal to the function $p \wedge (q \vee r)$?
Let $a(p,q,r)=(p \wedge q) \vee r$
$b(p,q,r)= p \wedge (q \vee r)$
By associate law $a=b$, but using $a(0,0,1)=1$ and $b(0,0,1)=0$?
Any suggestions?

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  • $\begingroup$ Welcome to MSE Asnil. I edited your question to use latex codes and I added the 'boolean algebra' tag. Please have a look so you know how to format question in the future. $\endgroup$ – Ittay Weiss Feb 26 '14 at 8:12
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The way the question is written it is not entirely clear what you are asking. You are asked whether two boolean expressions are equal. It seems that you are trying to use associativity somewhere, but it is not clear how (or why). The associative law absolutely does not say that $a=b$. I think that you are also trying to say that $a$ evaluated at $(0,0,1)$ is different than $b$ evaluation at the same values. That is reason enough to conclude the two expressions are different as functions.

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  • $\begingroup$ Associate law: (p∧q)∧r = p∧(q∧r) and (p∨q)∨r = p∨(q∨r). (p∧q)∨r = p∧(q∨r), here we changing brackets from p and q to q and r, am i doing it wrong because signs are different. $\endgroup$ – Asnil Deo Feb 26 '14 at 8:23
  • $\begingroup$ Firstly, i used associate law to confirm that a=b, which i assume is wrong. Secondly, i used a(0,0,1) and b(0,0,1) which gives two different answers. Since i wrongly used associate law the functions are not equal i guess. $\endgroup$ – Asnil Deo Feb 26 '14 at 8:29
  • $\begingroup$ you are correct! $\endgroup$ – Ittay Weiss Feb 26 '14 at 8:40

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