# Boolean algebra implementation question [closed]

We could use boolean algebra to analyse the digital circuit and use boolean simplification to optimise the circuit.

In boolean expression "1 + A" is 1, whatever A is 1 or 0.

But in the real digital world, the result would be 0 if expression is "1+1" and result is 1 bit. Does it mean we can't use the boolean algebra in any electronic case or I misunderstand something?

## closed as unclear what you're asking by Marcus Müller, Warren Hill, Finbarr, RoyC, BimpelrekkieApr 29 at 9:08

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• It's "boolean", like you've used it once in your question, not "boolen" like you used thrice :) No big deal, I fixed that for you. You're wrong however. You're confusing the "OR" with the "XOR" operation, and these are two different things. What "+" means depends on context, but if nothing else is said, I'd assume "OR" – Marcus Müller Apr 28 at 11:25
• As others have pointed out, boolean algebra chooses to define the meaning of the symbol $+$ to indicate "inclusive OR." (There are good reasons for the choice.) That algebra also chooses to define the meaning of the symbol $\oplus$ to indicate "exclusive OR." You need to avoid conflating the two into a mush. They mean different things. So you need to keep your head straight when reading forward. If someone writes "OR" without either "inclusive or exclusive" then you should probably assume a priority for $+$ ("inclusive OR"), unless the context makes it otherwise clear to be $\oplus$. – jonk Apr 28 at 16:55