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I've read the formula written on the title on a mathematics' book and it doesn't seem correct for me:

For the first part of the formula [A ∨ (B∧C)] I have the following possible values:

A; B and C; A and B and C (the Or is not exclusive)

For the second part of the formula [(A ∨ B) ∧ (A ∨ C)] I have the following values:

A and A (A); A and C; B and A; B and C; ...; A and B and A and C (A and B and C)

So I can have for the second part of the formula A and B; A and C which I can't obtain with the first part of the formula.

If I'm mistaken can somebody please tell me how and give some examples.

thanks,

Bruno

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    @Fredrik Meyer ∧ Alex Basson : Thanks you both for your answers ! Unfortunately I can accept only one. Good point for the most comprehensive one.2010-12-07

2 Answers 2

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Suppose you have $A \vee (B \wedge C)$. Then you either have $A$ or $B \wedge C$. If you have $A$, then you have $A \vee B$ and $A \vee C$. If, on the other hand, you have $B \wedge C$, then you have both $B$ and $C$, in which case you have both $A \vee B$ and $A \vee C$. So $A \vee (B \wedge C) \Rightarrow (A \vee B) \wedge (A \vee C)$.

Going in the other direction, suppose you have $(A \vee B) \wedge (A \vee C)$. So you have both $A \vee B$ and $A \vee C$. Either you have $A$, or you don't. If you have $A$, then you have $A \vee (B \wedge C)$. If, on the other hand, you don't have $A$, then the fact that you do have both $A \vee B$ and $A \vee C$ implies that you must have both $B$ and $C$. Therefore you have $B \wedge C$, and so you have $A \vee (B \wedge C)$, i.e. $(A \vee B) \wedge (A \vee C) \Rightarrow A \vee (B \wedge C)$.

So we've established that $A \vee (B \wedge C) \Rightarrow (A \vee B) \wedge (A \vee C)$ and $(A \vee B) \wedge (A \vee C) \Rightarrow A \vee (B \wedge C)$. Therefore $A \vee (B \wedge C) = (A \vee B) \wedge (A \vee C)$.

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Draw a Venn diagram and think of $A,B,C$ as sets. (consisting of True,False). Then you'll see that the formula is correct.

Or you could draw a truth table. http://en.wikipedia.org/wiki/Truth_table