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  #1  
Old 09-22-2010, 09:28 PM
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Question Boolean Algebra, equivalent expressions

Problem Statement: Using Boolean Algebra, determine whether or not the following expressions are valid:

\bar{x_{1}}x_{3} + x_{1}x_{2}\bar{x_{3}} + \bar{x_{1}}x_{2} + x_{1}\bar{x_{2}} = \bar{x_{2}}x_{3} + x_{1}\bar{x_{3}} + x_{2}\bar{x_{3}} + \bar{x_{1}}x_{2}x_{3}

I don't even know how to start questions like this. I'm not too bad in boolean algebra however when I'm doing a questions as complicated as this it's overwhelming.

I can't formulate an attack plan with an equivalent of this complexity, I feel like I just have to start adding terms that are equivalent to 0 and try to reduce to the RHS somehow.

How can I plan out my attack so that I have more of a chance of turning the LHS into the RHS? What specific characteristics or hints do you look for?

Thanks again!
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Old 09-22-2010, 10:43 PM
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Even though it's downright calculative, why don't you try to make the truth tables of the two expressions? They 're only 8 rows long and will definitely nail that problem.

However, if you want to stay true to the essense of the exercise and use boolean algebra, you can "disguise" a truth table into a sum of its terms, and then see wich terms appear on each side.

For example,
\bar{x_1}  x_3 \text{can be written}\\<br />
\bar{x_1} x_3 (x_2 + \bar{x_2})\\<br />
= \bar{x_1} x_3 x_2\ +\ \bar{x_1} x_3 \bar{x_2}

which clearly shows that this term contains minterms 3 and 1 (taking x3 as the Least Significant Bit).

I think this is the most strategic, fool-proof and thought-free attack plan you can use.
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Old 09-23-2010, 01:23 AM
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Question

Quote:
Originally Posted by Georacer View Post
Even though it's downright calculative, why don't you try to make the truth tables of the two expressions? They 're only 8 rows long and will definitely nail that problem.

However, if you want to stay true to the essense of the exercise and use boolean algebra, you can "disguise" a truth table into a sum of its terms, and then see wich terms appear on each side.

For example,
\bar{x_1} x_3 \text{can be written}\\<br />
\bar{x_1} x_3 (x_2 + \bar{x_2})\\<br />
= \bar{x_1} x_3 x_2\ +\ \bar{x_1} x_3 \bar{x_2}

which clearly shows that this term contains minterms 3 and 1 (taking x3 as the Least Significant Bit).

I think this is the most strategic, fool-proof and thought-free attack plan you can use.
Okay so here's my attempt at disguising a truth table in my boolean algebra expression for the LHS. I came up with the following,

\bar{x_{1}}x_{2}x_{3} + \bar{x_{1}}\bar{x_{2}}x_{3} + x_{1}x_{2}\bar{x_{3}} + \bar{x_{1}}x_{2}x_{3} + \bar{x_{1}}x_{2}\bar{x_{3}} + x_{1}\bar{x_{2}}x_{3} + x_{1}\bar{x_{2}}\bar{x_{3}}

With x3 being the least significant bit one can see that the following binary values will represent a 1 in our truth table,

011,
001,
110,
011 (repeated),
010,
101,
100.

I don't how I can use this information to turn the LHS into my RHS.

I could expand my RHS in terms of minterms as well and show that the truth tables are the same, but the questions asks me to use boolean algebra.

Any ideas?

EDIT: I figured it out. I do exactly what I stated above. Do what I just did to LHS to RHS now, and then I'll see that both terms contain the exact same terms, and therefore they are equivalent.

Last edited by jegues; 09-23-2010 at 03:44 AM.
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Old 09-23-2010, 12:01 PM
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You got it! And remember that duplicate terms can be simply ignored.
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Old 09-23-2010, 06:55 PM
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jegues jegues is offline
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Quote:
Originally Posted by Georacer View Post
You got it! And remember that duplicate terms can be simply ignored.
Right!

Because after all,

x + x = x
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