I have simplified following Boolean expressions. Can somebody tell me whether they are right or wrong?
1) F1 = ~(~A ~B C + ~(AB)C)
~(~A ~B C) = ~(~A) + ~(~B) + ~C -------> Apply DeMorgan's law to the 1st term
= A + B + ~C ---------> since A=~(~A)
~(~(AB) C) = ~(~A) + ~(~B) + ~C -------> Apply DeMorgan's law to the 2nd term
= A + B + ~C ---------> since A=~(~A)
Thus;
~(~A ~B C) + ~(~(AB) C) = (A + B + ~C) + (A + B + ~C)
Simplified expression = (A + B + ~C)
Regards
Your expression $$\neg\Big((\neg A \wedge \neg B \wedge C) \vee \big(\neg(A \wedge B) \wedge C\big)\Big)$$ is, by de Morgan,
$$\neg(\neg A \wedge \neg B \wedge C) \wedge \neg \big(\neg(A \wedge B) \wedge C\big).$$
By de Morgan again, this is
$$\big((A \vee B) \vee \neg C\big) \wedge \big((A \wedge B) \vee \neg C\big).$$
Now one wants to use distributivity. Since $\neg C$ occurs on both sides, the result will simplify to
$$\big((A \vee B) \wedge (A \wedge B)\big) \vee \neg C.$$
By distributivity and absorption, this is just $(A \wedge B) \vee \neg C$.