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Assume $A$, $B$, and $C$ are three independent predicates. Maybe $A$ stands for "my age is 20," and $B$ "stands for tomorrow is a good day."

So is it true that $(A \lor B) \land C \iff (A \lor C) \land (B \lor C)$?

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Short answer: No – user127.0.0.1 Jan 26 '14 at 15:10
No, it's (A and C) or (B and C). – TonyK Jan 26 '14 at 15:10
Ok , I made a mistake it should be is (A or B) and C equal to (A and C) or (B and C) – Simic Jan 26 '14 at 15:14
It is fine to change a question "midstream" after receiving answers. But when you do, and you've received answers, please scroll to the bottom of your post, add the word: EDIT, and then mention that you meant to post "x, y, and z". – amWhy Jan 26 '14 at 15:26
You can't simply pick values like this, it has to hold for all values. That is like claiming that $2x=x$ for all $x\in \mathbb{R}$ simply because $2.0=0$ – hmmmm Jan 26 '14 at 15:33

No, it's NOT true that $$(A \lor B) \land C \equiv (A \lor C) \land (B \lor C)\tag{NOT TRUE}$$

What is true, by the distributive law, is that $$(A\lor B)\land C \equiv (A\land C)\lor (B\land C)\tag{TRUE}$$

Confirm the correct equivalence below:

enter image description here

enter image description here

Compare with the following:

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EDIT: Note that there will be truth value assignments in which it turns out that the truth of $(A \lor B)\land C$ is equal to the truth value of $(A\lor C)\land (B\lor C)$ (and hence the resulting equal probabilities), (for example, see the first rows). But the two expressions are NOT the same under all possible truth value assignments (see for example the second rows, or the second to last rows), and therefore, are not equivalent.

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$ \begin{array}{|c|c|c|c|} A & B & C & (A \bigvee B) \bigwedge C \\ T & T & T & T\\ T & T & F & F \\ T & F & T &T \\ T & F & F & F \\ F & T & T & T \\ F & T & F & F \\ F & F & T & F \\ F & F & F & F \end{array}$

$ \begin{array}{|c|c|c|c|} A & B & C & (A \bigvee C) \bigwedge (B\bigvee C) \\ T & T & T & T\\ T & T & F & T \\ T & F & T &T \\ T & F & F & F \\ F & T & T & T \\ F & T & F & F \\ F & F & T & T \\ F & F & F & F \end{array}$

As the truth tables are not the same we can see they are not the same. It would have been easier just to try in on a few values till one didn't match but oh well!

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