The formula:

$r ∧ q ∧ ¬ p ∨ r ∧ p ∧ ¬ q ⇒ r ∨ ¬ p ∨ ¬ q$

The answer says it's a tautology but I've tried to work it out and I get false.

My truth table to the first half of the formula is in the photo and you can see that I got false so where have I gone wrong? Image

I've done the right hand side

enter image description here

So because the right side is true, it's a tautology?

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    $\begingroup$ please use parentheses. $\endgroup$ – Henno Brandsma May 12 '18 at 14:36
  • $\begingroup$ Welcome to MSE. Please use MathJax. $\endgroup$ – José Carlos Santos May 12 '18 at 14:36
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    $\begingroup$ @JoséCarlosSantos, please don't bite newcomers. $\endgroup$ – Henning Makholm May 12 '18 at 14:56
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    $\begingroup$ @JoséCarlosSantos: It is not a "suggestion" when you phase it like he should spend time learning a markup language just to please you. Learning MathJax is not a requirement for using this site. $\endgroup$ – Henning Makholm May 12 '18 at 15:00
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    $\begingroup$ Seeing all of the "Please use MathJax" comments, I am struck by how unwelcoming of a tone they tend to create. I started using MathJax because friendly people just edited it into my first few posts, and I caught on because it was obviously the thing to do. Be like those friendly people :) $\endgroup$ – G Tony Jacobs May 12 '18 at 16:29

In your truth table you only worked out the truth-conditions of $(r \land q \land \neg p) \lor (r \land p \land \neg q)$. Now you need to work out $r \lor \neg p \lor \neg q$, and then use the $\Rightarrow$ to get the truth-conditions of the whole conditional.

Remember that the $\Rightarrow$ is false when the antecedent is true and the consequent is false, and is true otherwise. So, in all those cases where $(r \land q \land \neg p) \lor (r \land p \land \neg q)$ evaluates to false, the whole conditional is automatically true, which covers $6$ out of the $8$ rows. So, you just need to check that in the other $2$ rows, $r \lor \neg p \lor \neg q$ is indeed true.

  • $\begingroup$ Oh, I thought both sides have to be true for it to be a tautology and because the left side is false, I didn't do the right side $\endgroup$ – Sian May 12 '18 at 14:42
  • $\begingroup$ @Sian No. When the left side (the 'antecedent') is false, the conditional is automatically true. $\endgroup$ – Bram28 May 12 '18 at 14:42
  • $\begingroup$ @Sian When the two "sides" are joined by $\land$ then both have to be true. That is the meaning of $\land$. But $\implies$ and $\iff$ do not work that way. False on the left of $\implies$ actually is good. What you have to worry about with $\implies$ is that it must not be false on the right and true on the left. $\endgroup$ – David K May 12 '18 at 14:46
  • $\begingroup$ Oh okay, I've done the right side, so would that be a tautology? $\endgroup$ – Sian May 12 '18 at 14:55
  • $\begingroup$ @Sian Like the left side, the right side by itself is not a tautology either. But when you put the $\Rightarrow$ between the left and right side, the whole thing becomes a tautology: there will never be a case where the left side is true and the right side false at the same time, and so the whole statement is never false, meaning it is a tautology. $\endgroup$ – Bram28 May 12 '18 at 14:57

You have gone wrong nowhere. It is a very common misconception among freshmen. An implication is false only when the antecedent is true and the consequent is false. For instance, the sentence "if it rains tomorrow, I'll carry an umbrella" is false only if it DOES rain and you DON'T carry an umbrella. If it doesn't rain, you've spoken truth, whether you carry an umbrella or not. It's also true if it does rain and, as a man or woman of your word, you do bring along an umbrella.

So, all those cases where you've gotten a false truth value for the antecedent? They straightaway mean that the full sentence is true! You need only examine whether the consequent is also true in those two other cases, and you'll have proven indeed that the above sentence is a tautology.

I'm sure you didn't doubt that anyway - after all, the antecedent implies that r is true, which is sufficient to deduce the consequent.

  • $\begingroup$ I'm understanding this so much better now, thank you $\endgroup$ – Sian May 12 '18 at 14:56

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