These are the 2 Morgan's laws:
1. NOT(A ^ B) = NOT(A) v NOT(B)
2. NOT(A v B) = NOT(A) ^ NOT(B)
This is the double negation law:
NOT(NOT(R)) = R
My solution (after 30 min) to find the second Morgan's law using the first Morgan's law and the double negation is this:
NOT(R) = NOT(A) v NOT(B)
NOT(NOT(A) v NOT(B)) = NOT(NOT(A ^ B)) = A ^ B
Do you think this is a valid demonstration?
Best Answer
You have $$\operatorname{NOT}(A \land B) = \operatorname{NOT}(A) \lor \operatorname{NOT}(B)\tag{T1}$$
Apply the NOT() to both sides like you did above:
$$\operatorname{NOT}(\operatorname{NOT}(A \land B)) = \operatorname{NOT}(\operatorname{NOT}(A) \lor \operatorname{NOT}(B)\tag{T2})$$
Apply double negation like you did above:
$$A \land B = \operatorname{NOT}(\operatorname{NOT}(A) \lor \operatorname{NOT}(B)\tag{T3})$$
Now let $X = \operatorname{NOT}(A)$ and $Y = \operatorname{NOT}(B)$, which by double negation gives
$$A = \operatorname{NOT}(X) \tag{T4}$$ $$B = \operatorname{NOT}(Y) \tag{T5}$$
So that (T3) becomes:
$$\operatorname{NOT}(X) \land \operatorname{NOT}(Y) = \operatorname{NOT}(X \lor Y\tag{T6})$$
And there you have the law again with the alternate conjunction/disjunction.