I am to determine if argument is valid by making truth table

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W= Warning lights will come on
P= Pressure is high
R=Relief valve is clogged

Then i have premises as

W $\leftrightarrows$ P AND R ,where the symbol indicates bi conditional statement (1st Premise)

Negatition R (2nd Premise)

Therefore W $\leftrightarrows$ P (Conclusion)

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Now i made truth table as usual with 8 rows and written all other stuff. In first row where W,P,R are false , the premises and conclusion is coming out to be true which makes argument valid. But textbook states that it is invalid. I would like to know where i am going wrong


  • $\begingroup$ As $R$ is false. "$P$ and $R$" is false too. And you get that $W$ is always false. $\endgroup$ – quid Jul 31 '15 at 15:33
  • $\begingroup$ Is there any reason why you are trying to use a truth table here? It doesn't seem at all necessary. $\endgroup$ – Daniel W. Farlow Jul 31 '15 at 15:40
  • $\begingroup$ @quid please check my work and let me know where is mistake $\endgroup$ – Taylor Ted Jul 31 '15 at 15:40

I don't see why you want to solve your problem using the truth table.

  • Firstly, here is an intuitive approach that I believe the first thing to do when you have such problems.

$$ ``\mbox{Warning lights is on}" \Longleftrightarrow ``\mbox{Pressure is high}" \wedge \ `` \mbox{Relief valve is clogged }" $$

is equivalent to :

$$ ``\mbox{Warning lights is } \color{#C00}{off} " \Longleftrightarrow `` \mbox{Pressure is } \color{#C00}{not} \mbox{ high}" \vee \ `` \mbox{Relief valve is } \color{#C00}{not} \mbox{ clogged }" \tag{P} $$

Now, we consider the statements $(P)$ to be true and that $``\mbox{the Relief valve is not clogged}"$, that implies that Warning lights is off ( regardless of the pressure). So the conclusion is obviously invalid (because the pressure can be too hight and the warning still off).

  • Now if the table is required by the question, you have to add an arrow in your table that contain the truth values of

$$ \left( (W \Longleftrightarrow R \wedge P) \wedge \lnot R \right) \Longrightarrow \left( W \Longleftrightarrow P \right) $$

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The claimed conclusion is not $W\leftrightarrow R$ as you wrote but rather it is $W\leftrightarrow P$. From the given $\neg R$ we can conclude that $P\land R$ is false and hence the equivalent $W$ is also false. If the claim were correct,we could infer $\neg P$, but we certainly can't.

Incidentally, your mistake in encoding the claim as $W\leftrightarrow R$ does give a valid conclusion: As seen above, $W$ is false and as both $W$ and $R$ are false, $W\leftrightarrow$ is true.

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  • $\begingroup$ no no it was a typo. i meant P and not R $\endgroup$ – Taylor Ted Jul 31 '15 at 15:35
  • $\begingroup$ I have uploaded my work, where is mistake? $\endgroup$ – Taylor Ted Jul 31 '15 at 15:40
  • $\begingroup$ The premises i have ticked are the ones should be consistent with each other, aren't they? $\endgroup$ – Taylor Ted Jul 31 '15 at 15:48
  • $\begingroup$ In the third line the premises $\neg R$ amd $ are both true, but the conlusion is false $\endgroup$ – Hagen von Eitzen Jul 31 '15 at 15:55
  • $\begingroup$ But i was talking about first line $\endgroup$ – Taylor Ted Jul 31 '15 at 15:56

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