I'm trying to simplify this expression.

a'b'c'+a'bc'+a'bc+ab'c+abc'+abc

so far I have

(a'b'c'+abc)+ (a'bc'+abc)+(a'bc+abc)+ (ab'c+abc)+ (abc'+abc)

b(a'+a)(c'+c) + bc(a'+a) + ac(b'+b) + ab(c'+c)

b + bc + ac + ab

I'm stuck here and the final solution is 3 terms with 5 literals.

Thank you for your help.

 

I'm stuck too, how did I pass my digital logic class? I did do a K-map and found the solution to be : = b + a'c' + ac

I need to refresh my boolean reduction, anyone help???

Steve

 

a'b'c'+a'bc' +a'bc +ab'c +abc'+abc

(next line, we know what is...)

a'(b'c'+bc'+bc)+a(b'c+bc'+bc)

(so, look for what is not!)

(abc'+a'bc)'

(b'(ac'+a'c))'

b + a'c + ac'

Make sense?

Scubasteve, I don't want to even tell you how long ago that was, and how I had to reach back for this one

I double-checked using a truth table
abc
000
001---
010
011
100---
101
110
111
The entries with the dashes were not in the original statement. Then I realized I'd messed it up the first time around LOL

One of my early work assignments was checking a 64Kbit Boolean table that had been burned into a EPROM by reading the waveform from a digital analyzer scope. That was a looooonnnnnggggg string of 1's and 0's. I thought my eyes were going to bleed to death by the time I was done.

 

hi Wookie,

Sadly enough, this was last semester for me! But, at least I got good at 5-bit K-maps

I think there is a problem with your solution, I checked it over, it doesn't seem to hold the same logic as the original problem.

I checked over = B + A'C' + AC and it holds. I didn't prove via boolean operations though, just via k-mapping it out..

Steve

Yours is B + A'C + AC'

abc                                =                        should be
000                                     0                          1
001              1                     1                           0 
010       1                            1                           1
011       1      1                     1                           1
100                      1              1                           0 
101                                      0                           1
110       1              1              1                           1
111       1                             1                            1