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Attaboy, Titanium; boolean algebra/DeMorgan's Theorem isn't just for gate-slinging hardware weenies. However, non-zero 'a' gives the true evaluation.
@king: a simplification often makes the original expression look silly, as in this case. There are cases, however, where the tested values might be coming from a port in inverted form or some other weird combination. Devising an exclusive-nor (if one doesn't have an exclusive-or instruction) can actually be done in fewer steps if one complicates the expression.
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