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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?

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

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  • \$\begingroup\$ 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" \$\endgroup\$ – Marcus Müller Apr 28 at 11:25
  • \$\begingroup\$ 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\$. \$\endgroup\$ – jonk Apr 28 at 16:55
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The + symbol is "OR" rather than "sum" when discussing boolean algebra.

The problem you are having is that you are confusing the usual meaning of + in mathematics with the boolean operator + for OR.

The function you describe is the boolean logic XOR (exlusive OR.)

This is also known as a half adder. 1 XOR 1 gives 0.

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