How do you prove $(a o+ b) o+ (~ a & ~ b) = 1$ ?

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Answer 1 · 2017-03-08T19:11:07

See explanation...

If I understand your notation correctly, it is what I would express as:

$(a vv b) vv (not a ^^ not b) = top$

If either of $a$ or $b$ is true, then $(a vv b)$ is true and hence:

$(a vv b) vv (not a ^^ not b) = top$

If neither of $a$ or $b$ is true, then $(not a ^^ not b)$ is true and hence:

$(a vv b) vv (not a ^^ not b) = top$

So under any combination of truth values for $a$ and $b$ , the given expression is true.

How do you prove (a o+ b) o+ (~ a & ~ b) = 1(a o+ b) o+ (~ a & ~ b) = 1 ?