My Math Forum Given truth table with 3 variables, find boolean expression

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 February 3rd, 2014, 10:45 AM #1 Senior Member   Joined: Mar 2012 Posts: 111 Thanks: 0 Given truth table with 3 variables, find boolean expression I don't know if I'm doing this right, but first I try to find the rule that most closely resembles the given problem. It's either XOR or XNOR. I chose XOR and created a blank truth table and tried to add combinations to XOR that would allow me to get rid of that one extra 1 (from XOR) and change the appropriate 1s to 0s and 0s to 1s. Nothing I've tried helps me solve this problem. I've tried (A?B?C).A' ; (A?B?C).(A.B') ; (A?B?C).(A'?C') ; and quite a few others. For this type of problem, are we supposed to just guess different combinations and just hope it all works out? Any help would be appreciated. Thanks!
 February 3rd, 2014, 12:50 PM #2 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 Re: Given truth table with 3 variables, find boolean express Just do it line by line. The last line has all three of A, B, and C being true, so it is A & B & C. The second-to-last line has C false, so it's A & B & -C. The combined expression so far is (A & B & C) | (A & B & -C). Just keep adding lines until you're done. (Of course you can go top to bottom as well.) Here I'm using & for AND, | for OR, and - for NOT; substitute whatever symbols your class uses.
February 3rd, 2014, 03:22 PM   #3
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Re: Given truth table with 3 variables, find boolean express

Hello, unwisetome3!

Quote:
 [color=beige]. . [/color]$\begin{array}{|ccc|c|} \;A\;&\;B\;&\;C\; & \;F\; \\ \\ \hline \\ \\ T&T&T&T \\ \\ T&T&F &F \\ \\ T&F&T & T \\ \\ T&F&F&T \\ \\ F&T&T& F \\ \\ F&T&F& F \\ \\ F&F&T & T \\ \\ F&F&F & T \\ \\ \hline \end{array}$

The statement is false in rows 2, 5, 6.

$(A\,\cap\,B\,\cap\,\bar{C})\,\cup\,(\bar{A}\,\cap\ ,B\,\cap\,C)\,\cup\,(\bar{A}\,\cap\,B\,\cap\,\bar{ C})$

$\;\;\;=\;B\,\cap\,\left[(A\,\cap\,\bar{C})\,\cup\,(\bar{A}\,\cap\,C)\,\cup \,(\bar{A}\,\cap\,\bar{C})\right]$

$\;\;\;=\;B\,\cap\,\left[(A\,\cap\,\bar{C}) \,\cup\{\bar{A}\,\cap\,(C\,\cup\,\bar{C})\}\right]$

$\;\;\;=\;B\,\cap\,\left[(A\,\cap\,\bar{C})\,\cup\,(\bar{A}\,\cap\,U)\right]$

$\;\;\;=\;B\,\cap\,\left[(A\,\cap\,\bar{C})\,\cup\,\bar{A}\right]$

$\;\;\;=\;B\,\cap\,\left[(A\,\cup\,\bar{A})\,\cap\,(\bar{C}\,\cup\,\bar{A}) \right]$

$\;\;\;=\;B\,\cap\,\left[U\,\cap\,(\bar{C}\,\cup\,\bar{A})\right]$

$\;\;\;=\;B\,\cap\,(\bar{C}\,\cup\,\bar{A})$

$\text{The statement is false for: }\:(\bar{A}\,\cup\,\bar{C})\,\cap\,B$

$\text{Therefore, it is true for:}$

$\;\;\;\bar{(\bar{A}\,\cup\,\bar{C})\,\cap\,B} \;=\; \bar{(\bar{A}\,\cup\,\bar{C})}\,\cup\,\bar{B} \;=\;(A\,\cap\,C)\,\cup\,\bar{B}$

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