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Validate or invalidate the following arguments

$ p\to t$

$ p \to \lnot r$

$q \to p$

$\lnot t \lor r$

$r \to t$

$\therefore \lnot p \land \lnot q \land (r \iff t) $

I could only see why it is $(r \iff t)$

LeonBrain
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1 Answers1

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Hint: You know $r \iff t$. What if $p$ were true?

Lord_Farin
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  • Thanks, I able to see the answer. But how do ermm, write the answer or prove it formally? – LeonBrain Nov 19 '13 at 09:10
  • That depends on the system you are using. This type of notation usually occurs in more or less informal settings. In that case, your proof could also be informal ("By $\neg t\lor r$ and $r\to t$, $r$ and $t$ are equivalent. If $p$ were true...") Or perhaps you need to use a truth table. Or you're given a proof system, with axioms and rules of inference -- this is the hardest option, since you need to formalise your argument using only the rules and axioms given, which may be much harder than the "intuitive" reasoning we used so far. – Lord_Farin Nov 19 '13 at 09:15
  • You can always use a truth table. – user99680 Nov 19 '13 at 09:16
  • @user99680 That is not true. You are taking for granted the soundness and completeness theorem for (a proof system for) propositional logic for granted here. These theorems require a decent amount of work. In any case, if OP has to prove it using a formal system, they cannot use a truth table. Simple as that. – Lord_Farin Nov 19 '13 at 09:19
  • @-Lord_Farin:I'm using the fact that propositional calculus is decidable. It doesn't seem that the OP's course would require to know the proof of this. IIRC the proof of decidability is not hard, given it is a finite system. – user99680 Nov 19 '13 at 09:29
  • Ultimately, you're dealing with a finite system; a finite collection of sentences, each with a well-defined truth value, and well-defined connectives, which are truth-preserving. – user99680 Nov 19 '13 at 09:34