Easy Bollean Algrebra

Thread Starter

smb3418

Joined May 5, 2007
2
Ok al I want to know is how this boolean expression is simplified to the final answer, I can prove it to myself in a truth table, but that takes too much time.

Here it is

x*y+not(x)*z+y*z

simplifies to
x*y+not(x)*z

Thanks

Scott

My first post
 

Salgat

Joined Dec 23, 2006
221
x*y+not(x)*z+y*z
=
x*y + z(not(x)+y)
x*y + z(not(x)+xy) - since not(x) + y = not(x) + x*y
x*y + z*not(x) + z*x*y
x*y + z*not(x)


I think thats right..?
 

BrianH

Joined Mar 21, 2007
43
Well, your answer is definitely good but I can't seem to follow your method. It's a pain that we can't write these things out more clearly I think. It'd be good if we could use a font where they had Not A's and NotB's etc in them.

I think the way to proceed with that problem is to use the Race Hazard Theorem. I think I'll take a look at this later on in the afternoon and see what I can come up with.

Brian
 

Salgat

Joined Dec 23, 2006
221
x*y+not(x)*z+y*z

=

x*y + z(not(x)+y)
- z is ANDed to both y and not(x), so you can put the two in parenthesis as being ORed

x*y + z(not(x)+x*y)
- since not(x) + y = not(x) + x*y (its a theorem)

x*y + z*not(x) + z*x*y
- converting it all back out by ANDing z to not(x) and z to x*y

x*y + z*not(x)
- since x*y already exists, the z in z*x*y becomes optional and can be removed
 

recca02

Joined Apr 2, 2007
1,214
- since not(x) + y = not(x) + x*y (its a theorem)
and here is the proof for that (since i cudnt digest it so i tried,my knowledge is basic and i dont remember much of digital electronics i learned myself for a brief period)

a' + ab = a'(1+b) +ab --------( 1+b)=1
= a' + a'b +ab
= a' + b(a' + a)
= a' + b(1)
= a' + b
 
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