Boolean algebra #24: DeMorgan's theorem - examples

[021radio]

thaks



[AZCHIM]

Thank you sir

I was confused on how the sing changes from . to + but I guess it's the rule :D

I thik I am clear now thank you so much . bless you XD

[AZCHIM]

Hello sir

I am confused why did you do (A+B+C) + D ? for the first question . Because After (A+B+C) it should be .D isn't it? but where dis "+" sing came before D like this " + D"?

thank you

[keleshevCOM]

@AZCHIM Hi, I'm not sure if I understood your question, but in the first example ((A+B+C)D)' where by (...)' I mean negated, we imagine that (A+B+C) is a first term in DeMorgans law, and D is the second. That's how we get (A+B+C)' + D'. it is the same as in (X * D)' = X' + D' but instead of X we have (A+B+C).

[SDTricker]

You're my hero. I am so going to pass this class now. Can't wait to see more.

[keleshevCOM]

@SDTricker Thanks! I'm so happy that this is useful :-) More stuff coming soon.