# Valid arguments and truth tables

I'm trying to understand validity of arguments and using truth tables. This concerns an example on a discrete math course on Linkedin Learning which I reproduce here. Essentially, trying to make sentences out of the truth table is not making much sense except for the first row.

We have some sentence propositions:

• Today is Monday ($$p$$)
• If today is Monday, then I will have a salad for lunch ($$p \rightarrow q$$)
• Therefore, I will have a salad for lunch ($$q$$)

Which, symbolically, looks like

$$p \\ p \rightarrow q \\ \therefore q$$

This is then used to construct the following table

$$\begin{array}{|c|c|c|} \hline p& p \rightarrow q & q \\ \hline T &\ \ T & T \\ \hline T & F & F \\ \hline F & T & T \\ \hline F & T & F \\ \hline \end{array}$$

I get how the columns for $$p$$ and $$q$$ are laid out but I don't understand how the true/false values for $$p \rightarrow q$$ for the 3rd, and 4th rows play out (and I'm not entirely sure about the second either):

• Second row: "Today is Monday. If it is a Monday, then I will not have a salad for lunch. Therefore I am not having a salad for lunch".
• Third row: "Today is not Monday. If it is Monday, then I will have a salad for lunch. Therefore I will have a salad for lunch" ?
• Fourth row: "Today is not Monday. If it is Monday, then I will have a salad for lunch. Therefore I will not have a salad for lunch" ?
• You have to check the def of Valid argument: applying it to truth table, you have to consider all the lines where both premise are TRUE. If in that lines the conclusion is also TRUE, then the argument is valid. Commented Jul 12, 2021 at 11:42
• As you can see, only line 1 has both premises: $p$ and $p \to q$ true. Commented Jul 12, 2021 at 11:43
• See similar post Commented Jul 12, 2021 at 12:52
• Thanks for the answers and comments all; I will soon read carefully read through them all and accept an answer. Commented Jul 13, 2021 at 5:44
• Dan's linked article is bogus. Commented Aug 1, 2021 at 10:42

## 3 Answers

You got the negations wrong. "If today is Monday, then I will have a salad for lunch" being false does not mean "If today is Monday, then I will not have a salad for lunch" (that would be $$p \to \neg q$$). It means "It is not the case that if today is Monday, then I will have a salad for lunch" ($$\neg (p \to q)$$). This is equivalent to saying "Today is Monday but I will not have a salad for lunch" ($$p \land \neg q$$).

Also, it is easier to think of $$p \to q$$ in terms of $$\neg p \lor q$$ (they are equivalent in classical logic). "If ... then ..." in the mathematical sense does not mean "If ... then ..." the way it is used in ordinary English, so don't overthink the "if ... then" thing. "If today is Monday, then I will have a salad for lunch" as a translation of $$p \to q$$ does not imply that there is any kind of causal relation between the antecedent and the consequent, it really just means "Today is not Monday or I will have salad for lunch".

So:

1. First row: "Today is Monday, and if it is Monday I will have salad for lunch. Therefore I will have salad for lunch." If $$p$$ and $$p \to q$$ are both true, then $$q$$ must be true as well.
2. Second row: "Today is Monday, and it is not the case that if it is Monday, I will have salad for lunch, i.e., it is Monday but I will not have salad for lunch. Therefore I will not have salad for lunch." $$p \to q$$ false is equivalent to $$p \land \neg q$$ being true, so in this case $$q$$ must be false.
3. Third and fourth row: "It is not Monday, and if it is Monday I will have salad for lunch, i.e. either it is not Monday or I will have salad for lunch. Therefore I may or may not have salad for lunch." Both $$q$$ being false and $$q$$ being true is consistent with $$p$$ and $$p \to q$$ true, that's why both combinations occur in the truth table.

To check validity of an argument, we need to make sure that the conclusion is true in every row where all premises are true. The only such row is the first row, about the last three rows we don't need to care. In the first row the conclusion is true as well, so truth is preserved in call cases, therefore the argument is valid.

• > "If ... then ..." in the mathematical sense does not mean "If ... then ..." the way it is used in ordinary English -- Yes, this is key to my confusion, I was attaching the ordinary meaning to these concepts but I'm slowly now coming to grips with the mathematical interpretation. Commented Jul 14, 2021 at 10:52

The “argument form” (“reasoning pattern”) in your example is called Modus Ponens. Notice that it has two premises, followed by a conclusion. To verify that the argument is valid is to verify that its associated conditional $$[p\;\&\;(p \rightarrow q)] \rightarrow q$$ is a tautology.

So we construct its truth table (order of construction: black, then cyan, then red):

$$\begin{array}{c|c|c|c} \text{premise 1} & &\text{premise 2} &\therefore &\text{conclusion} \\ [p &\& &(p \rightarrow q)] &\rightarrow & q \\\hline T &\color{cyan}T &\color{cyan}T &\color{red}T &T \\T &\color{cyan}F &\color{cyan}F &\color{red}T &F \\F &\color{cyan}F &\color{cyan}T &\color{red}T &T \\F &\color{cyan}F &\color{cyan}T &\color{red}T &F \\\end{array}$$

Since the entire column under the main connective (in red) is filled with just $$\color{red}T$$'s, said conditional in indeed a tautology, as desired.

The negation of "If it is Monday, then I will have a salad for lunch" $$(p\rightarrow q)$$ is actually "It is Monday and/but/yet I will not have a salad for lunch" $$(p\;\;\&\;\;\text{not }q)$$—not another conditional.

To be clear, let us first look at an invalid argument: $$p\land (p\lor q) \to q$$

$$\begin{array}{c|c|c|c|c} \text{p} &\text{q} &\text{Premise 1} &\text{Premise 2} &\text{Conclusion} \\~ & ~ &\text{p} & \text{p} \lor q & \text{q} \\\hline T &T &T &T &T \\T &F &T &T &F \\F &T &F &T &T \\F &F &F &F &F \\\end{array}$$

Here, we need to look only at the lines where both premises are true (only lines 1 and 2 here). On line 2, we see that the conclusion is false. Therefore, we have an invalid argument.

In your example we have : $$p\land (p\to q) \to q$$

$$\begin{array}{c|c|c|c|c} \text{p} &\text{q} &\text{Premise 1} &\text{Premise 2} &\text{Conclusion} \\~ & ~ &\text{p} & \text{p} \to q & \text{q} \\\hline T &T &T &T &T \\T &F &T &F &F \\F &T &F &T &T \\F &F &F &T &F \\\end{array}$$

Here, too, we need only look at the lines where both premises are true (only line 1 here). There, we see that the conclusion is true. Therefore, we have an valid argument.

You wrote:

We have some sentence propositions:

• Today is Monday ($$p$$)
• If today is Monday, then I will have a salad for lunch ($$p \rightarrow q$$)
• Therefore, I will have a salad for lunch ($$q$$)

You also seem a bit confused about precisely what we mean by the conditional statement in this case. We usually define:

$$p\to q~~ \equiv ~~ \neg(p \land \neg q)$$

(Though often given as a definition, we can also derive this result from other more fundamental rules of inference. See my blog posting on this topic here.)

So your Monday lunch conditional statement can be restated as:

It is not the case that it is Monday and I am not having salad for lunch.

So, if $$p$$ is false (i.e. it is not Monday) then the conditional statement in question will be true regardless of truth value of $$q$$ (i.e. whether or not you are have salad for lunch).