Constructing truth tables to determine the validity of a symbolic argument

Question

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

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 | Q | ¬Q | P ∨ ¬Q | P |

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

| T | T | F | T | T |

| T | F | T | T | T |

| F | T | F | F | T |

| F | F | T | T | F |

Row 2 shows all true premises and a true conclusion, however row three shows all true premises and a false conclusion (which would indicate invalidity).

I want to conclude that the presence of a row where a false conclusion is derived from true premises determines that the argument is invalid. However, I was just concerned that Row 2 (showing all true premises and a true conclusion) somehow somehow affect this.

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

Answer 1

Please note that your third column (the $\neg Q$ column) is just a helper column, and not an actual premise. You have only one premise here, and that is the fourth column. So, you actually have three rows where the premise is true:

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

But yes, since in the last row the premise is true, but the conclusion is false, the argument is invalid. The fact that you have other rows where the premise is true and the conclusion is true as well does not take away from this.

Answer 2

The third row is wrong, because you have to consider P false and Q true.

The implication is false only when the hypothesis is false and the thesis is true. In the other cases is always true.

The final table is this:

| T | T | F | T | T |

| T | F | T | T | T |

| F | T | F | F | F |

| F | F | T | T | T | contradiction because the starting P is false, but after the implication it’s true.