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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 → ¬P) → P) → P

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

I have used these column headers:

| P | ¬P | P → ¬P | (P → ¬P) → P | P |

After filling out this truth table I have found that the first row reads:

| T | F | F | T | T |

And the second row reads:

| F | T | T | F | T |

Neither rows show all true premises and a true conclusion, however neither show all true premises and a false conclusion (which would indicate invalidity).

However I am not sure whether the absence of a row where there a false conclusion from true premises allows me to confidently read that the argument is a valid one.

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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Here's a tidied up table (you really should learn Mathjax, by the way):

\begin{array}{c|c|c|c|c} P&\neg P & P \rightarrow \neg P & (P \rightarrow \neg P) \rightarrow P&P\\ \hline T&F&F&T&T\\ F&T&T&F&F\\ \end{array}

The premise is the fourth column, and the conclusion is the fifth (I am assuming that is how to read your exercise, though maybe this is one of those arguments where there is no premise and the whole statement is the conclusion? That is, maybe the question is whether or not this whole statement is valid (a tautology)? I'll leave that up to you to tease out)

There is no row where the premise is true and the conclusion is false, and so this is a valid argument.

Contrary to what you say, though, there is a row where both the premise is true and the conclusion is true. I think that, as in your previous question, you are treating some of the helper columns as premises ... please differentiate between the two!

Anyway, having a row with a true premise and a true conclusion does not mean the argument is valid, as that only shows that it is possible for the conclusion to be true if the premise is true, and what validity needs is that the conclusion is necessarily true (i.e always true) whenever the premise is true.

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Your table should read: $$\begin{array}{c|c:c:c|c} P&\neg P & P \to \neg P & (P \to \neg P) \to P&((P\to\neg P)\to P)\to P\\ \hline T&F&F&T&T\\ F&T&T&F&T\\ \end{array}$$

Neither rows show all true premises and a true conclusion, however neither show all true premises and a false conclusion (which would indicate invalidity).

They do not need to.   A tautology is a statement that evaluates to true for all possible truth assignments of the literals.

The final column is true for both assignments of the only literal (first column) and so therefore that statement is a tautology.

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To show a classical proposition is valid with truth tables, you consider every way of assigning true and false to the propositional variables, and then check that the proposition is true for each assignment. In this case, there is only one propositional variable, namely P. Your table is only going to have two columns: P and ((P → ¬P) → P) → P. The first column will contain a row for true and a row for false, and the question is whether the second column contains true for both of those rows. If it does, then the proposition is a classical tautology, otherwise it isn't. There's no need to have columns for subformulas; their values are determined by the values assigned to the propositional variables.

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