cant understand boolean algebra for the life of me!

Thread Starter

Maxoriley618

Joined Sep 9, 2012
13
Determine whether or not the following expressions are valid, i.e., whether the left- and
right-hand sides represent same function.
(a) x1'x3 + x1x2x3' + x1'x2 + x1x2' = x2'x3 + x1x3' + x2x3' + x1'x2x3
(b) x1x3' + x2x3 + x2'x3' = (x1 + x2' + x3)(x1 + x2 + x3')(x1' + x2 + x3')
(c) (x1 + x3)(x1' + x2' + x3')(x1' + x2) = (x1 + x2)(x2 + x3)(x1' + x3')

the book that i have is on PDF and shows very little on what to do it just gives the theorems and example problems that have steps where they apply 4 theorems at once and an end result with no work shown. I'm not looking for simply answers (thought I'm guessing there are forum rules against it) but a detailed explanation of it would be wonderful.
 

Papabravo

Joined Feb 24, 2006
15,525
You can build a truth table.

Each of the terms in the expression represents one or more lines in the truth table. Write a 1 in the output column for each term. If the two truth tables are the same the expressions are equivalent.

LHS of the first expression
X1'X3 --> 0x1 --> 001, 011
X1X2X3 --> 111
X1'X2 --> 01x --> 010, 011
X1X2' --> 10x --> 100, 101

The 1's in the truth table are on lines 1,2,3,4,5,7

Rich (BB code):
X1 X2 X3 | Y
---------------
0  0  0  | 0
0  0  1  | 1
0  1  0  | 1
0  1  1  | 1
1  0  0  | 1
1  0  1  | 1
1  1  0  | 0
1  1  1  | 1
Get the idea?
 

MrChips

Joined Oct 2, 2009
23,090
Nice. Your example is a lot clearer than my explanation.
You're just a bit late joining in the party. I think they got it and went out for a beer.
 

WBahn

Joined Mar 31, 2012
26,398
The problems very likely expect solutions using Boolean algebraic manipulation. Using a truth table is a wonderful check (and it can also offer insight into the Boolean algebra that needs to be performed), but it may not be acceptable as the solution to the homework.

On parts (b) and (c), I would recommend walking a truth table across for the expressions as given, then multiply out the expressions on each side to yield a sum of products for each side. It's not as bad as it might seem at first because lots of the products will be eliminated and or redundant. Then work with what's left.
 
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