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September 15th, 2014, 06:41 AM   #11
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The fail-proof way to show that two Boolean expressions are equivalent is to construct truth tables for both of them. If the truth tables are the same, then they are equivalent.

EDIT: Sorry, missed the part about doing it algebraically.
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September 15th, 2014, 07:08 AM   #12
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Originally Posted by Sokrates View Post
Hm? In the task it says "Show (algebraically, using the that the laws that I wrote in the first post) that the following function expressions f1 to f4 are equivalent". How is that possible then?
If they're different it's not possible. Trick question, typo in the problem, you typed it out wrong, or I made a mistake -- take your pick. I did give a way to verify that the two are different, and you should check that to your satisfaction.
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September 17th, 2014, 09:29 AM   #13
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I discovered a solution, so I thought that I would share it here, maybe someone else will also find it useful.

Starting with $f_{3} = (a \vee c')(b' \vee c')(a' \vee d)$

Using the distributive law
$a\vee (bc)=(a\vee b)(a\vee c)$

We get $(a \vee c')(b' \vee c')(a' \vee d)= (c' \vee ab')(a' \vee d)$

Then multiplying out the parantheses, we get

$(c' \vee ab')(a' \vee d)=a'c' \vee c'd \vee aa'b' \vee ab'd$

In the above expression we have the term $aa'b'$ and since $aa'=0$ it can be removed. So then we have left

$a'c' \vee c'd \vee ab'd$

Which equals $f1$.


Last edited by Sokrates; September 17th, 2014 at 09:35 AM.
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September 18th, 2014, 11:47 AM   #14
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So f3 and f1 are not the same
CRGreathouse: Care to comment?

Last edited by Sokrates; September 18th, 2014 at 11:59 AM.
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September 18th, 2014, 12:55 PM   #15
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CRGreathouse: Care to comment?
I gave conditions under which f1 and f3 differ. Anyone interested can check the functions at those points. If the functions are the same at those points then I was mistaken (in proof if not conclusion). If they differ then the simplification is wrong. It doesn't mean much to me either way, this is just a 1-off problem.
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