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Proving Validity of a Symbolic Argument Using Truth Tables

I am looking to determine the validity of this argument using the truth table method:

P ∨ Q, P → R, ¬R ⊨ Q

I cannot yet embed the image of my truth table on here because I haven't earned emough points of my profile yet, so I will try to explain it the best I can.

I have used the column headers | P | Q | R |¬R | P ∨ Q | P → R | Q |

After filling out this truth table I have found that in row 3, the premises 'P ∨ Q' and 'P → R' are true and yet the conclusion is false, which would indicate that the argument is invalid.

However I am wondering whether I should also take into account the truth value of the premise '¬R'. If I do this, then I find only one critical row (row 6), and here I find the conclusion to be true. As there are no rows where the premises are true and the conclusion is false, I read from this that the argument is valid.

However, I am apprehensive to draw this conclusion; for one because I am not sure whether I have used the correct column headers, and for another because I am not sure whether I should be be looking at '¬R' as a separate premise and considering it's truth value to determine the validity of the conclusion.

As you can probably tell, I am a beginner in logic, so I would appreciate any help to clarify this. Thank you.

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  • $\begingroup$ See wolframalpha.com/input/… $\endgroup$ – Dan Christensen Nov 29 '17 at 19:08
  • $\begingroup$ @Dan Christensen Do you really think that WA is the absolute answer for this question ? $\endgroup$ – Jean Marie Nov 29 '17 at 19:14
  • $\begingroup$ You have to connect your different expressions by "&" connector, i.e., build the truth table of $(P ∨ Q)&(P → R)&(¬R)→Q$ and verify it's a tautology. Remark: Mapping ⊨ onto → is the only thing you can do in the context of boolean logic. $\endgroup$ – Jean Marie Nov 29 '17 at 19:20
  • $\begingroup$ @JeanMarie Thank you for your advice, that is really helpful. $\endgroup$ – user508231 Nov 29 '17 at 19:48
  • $\begingroup$ A truth table is more bother than it is worth when there is a simple proof of the statement. $\endgroup$ – William Elliot Nov 30 '17 at 2:24
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An argument is valid if and only if it is impossible for the conclusion to be false while all of its premises are true.

So yes, you do have to take into account all of the premises, including $\neg R$. And since you found that in the only row where all of the premises are true, the conclusion is also true, the argument is valid.

P.s. Here is how you can make a truth-table using Mathjax:

\begin{array}{ccc|c|c|c|c} P&Q&R&P \lor Q & P \rightarrow R & \neg R & Q\\ \hline T&T&T&T&T&F&T\\ T&T&F&T&F&T&T\\ T&F&T&T&T&F&F\\ T&F&F&T&F&T&F\\ F&T&T&T&T&F&T\\ F&T&F&T&T&T&T\\ F&F&T&F&T&F&F\\ F&F&F&F&T&T&F\\ \end{array}

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