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Is it legal to distribute the $\land$ or the $\lor$ operators over the $\implies$ operator? For example is it legal for me to to do the following?

$(p\land(p\implies q))\implies q $

$(p\land p) \implies (p\land q) \implies q$

Put more simply: Is the "AND" operator distributive over the "IMPLIES" operator.

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It is legal in the U.S. in the sense that there is no law that requires you to do only logically correct things, so writing this will not break any law (and I'm fairly sure that this is also true elsewhere in the world). However, this is not a rule of inference that is a tautology. –  Arturo Magidin Jan 15 '11 at 0:19

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No, in general you can't replace $a \wedge (b \implies c)$ with $(a \wedge b) \implies (a \wedge c)$. If $a$ is true, then both expressions have the same truth value. However, if $a$ is false, then the first expression is false and the second expression is true.

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The "truth table" http://en.wikipedia.org/wiki/Truth_table is an easy method to verify and even proove such assumptions.

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Thanks for the link, but I'm not trying to prove this proposition. I'm trying to find out if the "AND" operator is distributive over the "IMPLIES" operator. –  lampShade Jan 14 '11 at 20:48
2  
Then build a truth table for both equations and see if its identical? –  Listing Jan 14 '11 at 20:54

Assuming that $\implies$ distributes to the right (so by $A \implies B \implies C$ you mean $A \implies (B \implies C)$), then for your examples it works out, since they're both tautologies.

Otherwise, your example doesn't work it, since the second statement is no longer a tautology.

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