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Let $p,q$ be elementry statements and $\alpha,\beta,\gamma$ be statements. (sorry if this is the wrong translation).

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Prove/disprove:

  1. is $p,q\Rightarrow \gamma$ tautology?

  2. is $\alpha,\beta\Rightarrow \gamma$ tautology?

I think both are wrong since we have no information about the part in the red square below, for all we know $\gamma$ can be false with everything else. But it seems too easy so I'm doubting myself..

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Note: $\Rightarrow$ means tautology, comma means $\wedge$.

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  • $\begingroup$ I think you're missing a bit of the problem. I'm assuming it should be something like "let $\alpha, \beta, \gamma$ be statements that comply with the first row of the truth table below", otherwise the questions are trivial. Edit: With my interpretation of the problem, you don't need the other rows of the truth table to answer the first question.You're only interested on when the premisses are both true and you have that information. $\endgroup$ – Git Gud Jan 17 '15 at 11:22
  • $\begingroup$ @GitGud it doesn't say that exactly but yeah they have to comply with the first row. $\endgroup$ – shinzou Jan 17 '15 at 11:24
  • $\begingroup$ @GitGud doesn't $\gamma$ have to be true for any placing of true or false in $p,q$ for it to be tautology? $\endgroup$ – shinzou Jan 17 '15 at 11:27
  • $\begingroup$ No, it need only be true when $p$ and $q$ are. I'm not exactly sure of the meaning you're giving to $\color{red},$ and $\color{red}\implies$, but whatever (reasonable) meaning it is, the idea behind them is the same. It's an argument of the form $A,\text{ therefore }B$ and you're being asked to say when is such an argument valid. I think you know how to answer this and that you can make parallel between this and the problem. $\endgroup$ – Git Gud Jan 17 '15 at 11:30
  • $\begingroup$ I deleted my previous comment because I misread something and as a consequence said wrote wrong. To prove that the second one isn't a tautology you need to find $\alpha,\beta,\gamma$ such that there is a row of the truth table with the property that $\alpha$ and $\beta$ are true, but $\gamma$ is false. Hint to find appropriate formulas: just let $\alpha$ and $\beta$ be tautologies. Edit: Answering the question in your previous comment, your choices do not help achieving what I described in the present comment. $\endgroup$ – Git Gud Jan 17 '15 at 11:57
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  1. True, since by the definition of tautology:

From wiki

A formula of propositional logic is a tautology if the formula itself is always true regardless of which valuation is used for the propositional variables

So we can see from the truth table that $p,q,\gamma$ are all true therefore it's a tautology.

  1. False, take $\alpha=p\vee\neg p,\beta=q\vee\neg q, \gamma=(p\wedge q)$ so for $p=0,q=0$, $\alpha=1,\beta=1$ but $\gamma=0$ therefore it isn't tautology.

Thanks to Git Gud for the help.

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    $\begingroup$ Your definition of tautology is wrong, a tautology is a statement that is true regardless of the truth values you give to its components. In the case of a conditional statement, to see that it is a tautology, it suffices to check that the consequent is true whenever the antecedent is (why does this suffice?). This is something that the truth table tells us it happens, so you can conclude it is a tautology. Please fix your answer and most importantly understand this. $\endgroup$ – Git Gud Jan 17 '15 at 12:16
  • $\begingroup$ @GitGud I actually copied it from: mathworld.wolfram.com/Tautology.html It suffices because with anything else, the premises can be false so the consequence would be trivially true? $\endgroup$ – shinzou Jan 17 '15 at 12:39

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