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September 15th, 2014, 06:41 AM  #11 
Member Joined: Aug 2014 From: Somewhere between order and chaos Posts: 44 Thanks: 5 Math Focus: Set theory, abstract algebra, analysis, computer science 
The failproof 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. 
September 15th, 2014, 07:08 AM  #12 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  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.

September 17th, 2014, 09:29 AM  #13 
Newbie Joined: Sep 2014 From: Ancient Greece Posts: 10 Thanks: 0 
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. 
September 18th, 2014, 11:47 AM  #14 
Newbie Joined: Sep 2014 From: Ancient Greece Posts: 10 Thanks: 0  CRGreathouse: Care to comment? Last edited by Sokrates; September 18th, 2014 at 11:59 AM. 
September 18th, 2014, 12:55 PM  #15 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  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 1off problem.


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