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I have a question that I am stuck on reasoning about knowledge. For the following statement, I need to show whether it is valid, satisfiable or unsatisfiable:

$\bf{K}$(Happy $\lor$Sad) $\implies$ $\neg \bf{K}($Happy)

I have done the following:

Let e w be an interpretation where e is the set of possible worlds and w is the real world. Then we need to show that for any interpretation, e w satisfies $\neg \bf{K}$(Happy $\lor$Sad) $\lor$ $\neg \bf{K}$(Happy).:

Now assume that such an interpretation e w satisfies $\bf{K}$(Happy $\lor$Sad). This would mean for all w' $ \in$ e, e w' satisfies $\bf{K}$(Happy $\lor$Sad). However in a world w'' = {$\neg Happy, \neg Sad$}, then the interpretation e w'' does not satisfy $\bf{K}$(Happy $\lor$ Sad).

Therefore, we have a contradiction and for any interpretation e w, e w satisfies $\neg \bf{K}$(Happy $\lor$Sad).

Similarly, assume an interpretation e w satisfies $\bf{K}($Happy). However, if we have a world(s) where $\neg$Happy is true, then e w'' does not satisfy $\bf{K}($Happy). Therefore for any e w, e w satisfies $\neg \bf{K}($Happy) and thus the statement is valid.

I'm not sure whether my proof is correct and if not, could give me a hint on how to go about with this question.

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1 Answer 1

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The formula is either satisfiable or unsatisfiable. Only when it is satisfiable, it may also be valid.

To rule out that the implication is valid, show that its negation is satisfiable, that is, construct a model that satisfies the LHS but not the RHS. (Hint: a model with just one possible world suffices.)

Next, rule out that the implication is unsatisfiable by providing a model that satisfies the RHS. (Hint: again, a model with just one possible world will do.)

After these two steps, you're left with the answer.

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    $\begingroup$ Why do we need two possible world to rule out the implication is unsatisfiable. Isn't a model with one world sufficient? $\endgroup$
    – user24376
    Sep 28, 2018 at 3:00
  • $\begingroup$ @user24376 Absolutely. (So my hint is making it more difficult than necessary.) $\endgroup$
    – Kai
    Sep 28, 2018 at 13:07

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