A little electronics question about logical gates

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Femme_physics said:
Isn't that the definition of ##\overline { \overline{ab}c }## ?

No you can't write it like that but you can use de morgan law to solve that:

[itex]\overline{\overline{ab}c}[/itex]

[itex]\overline{\overline{ab}} + \overline{c}[/itex]

[itex]ab + \overline{c}[/itex]

But using: [itex]\overline{ab} = \overline{a} + \overline{b}[/itex]
 


Looks like you got the hang of boolean algebra! :approve:

However, you dropped something when you had ##\overline{cc}## and went to ##cc##.EDIT: I missed the abc part which is not right. Fixed it in post #38.
 
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I like Serena said:
Looks like you got the hang of boolean algebra! :approve:

However, you dropped something when you had ##\overline{cc}## and went to ##cc##.

True, but at the end of the day the result is still 0 cause we got a and a(capped) in the same multiplication line so everything gets nullified, right?
 


Femme_physics said:
True, but at the end of the day the result is still 0 cause we got a and a(capped) in the same multiplication line so everything gets nullified, right?

Actually, you don't have ##\overline{a}##, but you have ##\overline{ab}##.
But yes, in combination with ##ab##, everything gets nullified.
 


I like Serena said:
Actually, you don't have ##\overline{a}##, but you have ##\overline{ab}##.
But yes, in combination with ##ab##, everything gets nullified.

Ah, gotcha! Now I am asked to write the logical value of the function Out, when the values of all the ports
C = '0'
A = B = '1'

But still, everything gets nullified right?
 

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