Boolean Algebra - Simplification

Thread Starter

alb_ross

Joined Jul 21, 2018
1
Hi All

I've seen multiple threads on this, but I'm having a bit of trouble remembering / getting back into the topic of boolean algebra. After taking note of all the essential rules, I'm not trying to simplify some equations I've found online.

This one seems extremely easy, but I just can't seem to wrap my mind around the boolean algebra identities to solve this:
(xy)+(xz)+(yz')

Solution for simplification is supposed to be:
(xz)+(yz')

I've tried to use the distributive and absorption law, but all I get is getting back to the original equation...

Could you give me a hint?

Thank you in advance!
 

MrAl

Joined Jun 17, 2014
7,757
Hi,

Also interesting is if you take the two term solution and invert it and simplify, then invert that also and simplify, you get the original three term expression which includes the term xy which shows that the term xy is superfluous.
 

MrChips

Joined Oct 2, 2009
21,665
Expand each of the original three terms:

xy = xyz + xyz'
xz = xyz + xy'z
yz' = xyz' + x'yz'

You will observe that the expansion of (xy) is repeated in the other two expansions.

As a check, draw the Karnaugh map and then it will become obvious that (xy) is redundant.
However, it becomes important if one needed to prevent glitches in an encoder circuit.
 

MrAl

Joined Jun 17, 2014
7,757
Expand each of the original three terms:

xy = xyz + xyz'
xz = xyz + xy'z
yz' = xyz' + x'yz'

You will observe that the expansion of (xy) is repeated in the other two expansions.

As a check, draw the Karnaugh map and then it will become obvious that (xy) is redundant.
However, it becomes important if one needed to prevent glitches in an encoder circuit.
Hi,

Could you elaborate a little with that?
I just woke up so maybe that's why i am not following your logic :)
 

MrChips

Joined Oct 2, 2009
21,665
xy can be expanded into xyz + xyz'.

xyz is also contained in xz.
xyz' is also contained in yz'.

Hence we can eliminate xy.

Summary here:
xy = xyz + xyz'
xz = xyz + xy'z
yz' = xyz' + x'yz'
 
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