# Simplify Boolean equations

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

• wolframalpha.com comes up with a different result for F1. F2 and F3 are looking ok. – Axel Kemper Jun 11 '14 at 16:24
• can you mention the answer you come up with F1? – user156436 Jun 12 '14 at 1:58
• It is the same solution as derived by jdc. – Axel Kemper Jun 12 '14 at 6:37

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 $(A \wedge B) \vee \neg C$.

• Accept solution? – jdc Jun 12 '14 at 18:04
• yes... I checked it and it is correct – user156436 Jun 13 '14 at 4:58
• There's a check mark one can select to award a responder "points." My hope is that if I accumulate enough of these points, people here will take me seriously enough that they will answer the questions that I ask. – jdc Jun 13 '14 at 7:21