Is
f = a'ab;
the same as
f = 1
Is this possible? I got this when I was simplifying something.
If that's meant to represent (and I think this is the likely case):
NOT-a AND a AND b
then, no, it's false no matter the values of a
or b
. That's because one of a
or NOT-a
is definitely false, FALSE AND anything
is false and the operation is associative: (a AND b) AND c == a AND (b AND c)
.
If it's meant to represent:
NOT-a OR a OR b
then, yes, it's true no matter the values of a
or b
. That's because one of a
or NOT-a
is definitely true and TRUE OR anything
is true. The associativity rules also apply here.