Is
f = a'ab;
the same as
f = 1
Is this possible? I got this when I was simplifying something.
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If that’s meant to represent (and I think this is the likely case):
then, no, it’s false no matter the values of
aorb. That’s because one ofaorNOT-ais definitely false,FALSE AND anythingis false and the operation is associative:(a AND b) AND c == a AND (b AND c).If it’s meant to represent:
then, yes, it’s true no matter the values of
aorb. That’s because one ofaorNOT-ais definitely true andTRUE OR anythingis true. The associativity rules also apply here.