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tyteen4a03
 one year ago
I've got (((P → Q) ∧ (R → S) ∧ (¬Q ∨ ¬S)) → (¬P ∨ ¬R)). Are there any quick and dirty ways to prove that it's a tautology/contradiction/neither without drawing a huge truth table?
tyteen4a03
 one year ago
I've got (((P → Q) ∧ (R → S) ∧ (¬Q ∨ ¬S)) → (¬P ∨ ¬R)). Are there any quick and dirty ways to prove that it's a tautology/contradiction/neither without drawing a huge truth table?

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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2hmm, looks like this expression is true after all.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2I haven't studied boolean algebra/formal logic so here's layman's perspective on what this expression is really trying to say. We don't care about what happens if the lefthandside turns out false because then the whole expression is true anyway but we're interested in the case where the lefthandside is true. So suppose that one of \(Q\) or \(S\) is false (\(\neg Q \wedge \neg S\)). The only way, then, that this lefthandside can be true is when one of \(P\) or \(Q\) is false, because implication means that if the righthandside is false, then the left has to be false too! And that is indeed the righthandside of the whole expression. :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(((P → Q) ∧ (R → S) ∧ (¬Q ∨ ¬S)) → (¬P ∨ ¬R)) assume P > Q, R > S, ~Q  ~S. now consider if ~Q. it follows then by contraposition that P > Q is equivalent to ~Q > ~P and so we can conclude ~P. otherwise, consider if ~S. similarly, by contraposition we have that R > S is equivalent to ~S > ~R so it follows that we can conclude ~R. so it follows that we can conclude ~P  ~R.
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