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I've found Isn't the modus ponens just the definition of what 'if' means? but I don't feel like it is particularly relevant. I also found Why is modus ponens a q-implication? which I understand if I don't dig too deep ($p$'s truth and $q$'s truth) but I don't feel like I fully understand modus ponents.

Modus ponens states

$$ p \rightarrow q $$

Given $p$

Therefore $q$

I'm trying to approach this by looking at the truth table but it isn't helping me.

$p$ $q$ $p\rightarrow q$
1 1 1
1 0 0
0 1 1
0 0 1

What is modus ponens saying exactly? I'm having a problem interpreting it.

Is it saying that given some $p$ (can this be true or false or must this only be true) and given some implication (can THIS be true or false or must this be true)?

We can determine that $q$ is true?

Assuming $p$ must be true and $p\rightarrow q$ must be true that DOES imply that $q$ must be true but in any other case we cannot determine $q$?

jMdA
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IanQ
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    Look at this as a mathematical equivalent of a hypothetical. "If hypothetically I was born in 2000, then I would be in my twenties this year." This is a true statement. But I was not born in 2000. But it's still possible that I could be in my twenties, or not. (In terms of mathematics, both $q$ and $\neg q$ are consistent with $p\rightarrow q$ and $\neg p$.) – jMdA Feb 02 '21 at 06:04
  • Q1) Sorry, I don't quite follow... If I understand what you said from " "If hypothetically I was born in 2000, then I would be in my twenties this year." This is a true statement", we must have (p->q) be true? – IanQ Feb 02 '21 at 06:17
  • Q2) "But I was not born in 2000. But it's still possible that I could be in my twenties, or not" this corresponds to p possibly being false but (p->q) still being true? – IanQ Feb 02 '21 at 06:18
  • Q3) "In terms of mathematics, both q and ¬q are consistent with p→q and ¬p." Sorry, I don't follow. What does consistent mean here? – IanQ Feb 02 '21 at 06:21
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    Q1) Yes, think about it, $p\rightarrow q$ is true in this case. – jMdA Feb 02 '21 at 06:22
  • Q2) Indeed, the point is that $p\rightarrow q$ is true, and this is independent from $p$ being true or not. – jMdA Feb 02 '21 at 06:23
  • Q3) consistent means, it doesn't cause a contradiction. $p\rightarrow q$ and $\neg p$ and $q$ can be true at the same time. Similarly $p \rightarrow q$ and $\neg p$ and $\neg q$ can be true at the same time. – jMdA Feb 02 '21 at 06:24
  • Let us continue this discussion in chat. – IanQ Feb 02 '21 at 06:32

1 Answers1

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Abstract answer. An inference rule is something of the form (for some $n \geq 0$)

$$\tag{*} \frac{H_1 \quad \dots \quad H_n}{T} $$

where $H_1, \dots, H_n, T$ are formulas; more precisely, $H_1, \dots, H_n$ are the hypotheses or premises of $(*)$ and $T$ is the thesis or conclusion of $(*)$.

The inference rule $(*)$ is read as follows: "Given $H_1, \dots, H_n$, then $T$ follows" (or similar expressions). This means that whenever $H_1, \dots, H_n$ hold simultaneously true, then by necessity $T$ holds true too. Said differently, it is impossible that $H_1, \dots, H_n$ are all simultaneously true and $T$ is not.

Concrete answer. Modus ponens is the following inference rule:

$$ \frac{p \to q \qquad p}{q} $$

According to the abstract explanation above, modus ponens says that whenever $p \to q$ and $p$ are simultaneously true, then $q$ is necessarily true. Indeed, this is exactly what happens in the truth table you wrote. Check the rows where both $p\to q$ and $p$ are simultaneously true (this happens only on the first row): in each of them (i.e. in the first row) it turns out that $q$ is true too. Said differently, in the truth table you wrote, it is impossible to find a row where $p \to q$ and $p$ are both true and $q$ is false.

Taroccoesbrocco
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  • Thank you very much! It wasn't clear to me whether (p->q) and p had to have concrete values (true) and thus I was not clear on what we could say about q. I understand now that inference rules must be true. Thus, given some p which is true, we can say something about q? – IanQ Feb 02 '21 at 06:57
  • This doesn't explain why $p \to q$ is defined the way it is for $(p,q) \neq (1,1)$. – Jordan Mitchell Barrett Feb 02 '21 at 06:58
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    @JordanMitchellBarrett - The OP question is about the inference rule called modus ponens, not about the meaning of the formula $p \to q$ (given by the truth table). – Taroccoesbrocco Feb 02 '21 at 07:03
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    @user162264 - Modus ponens says that whenever $p$ and $p \to q$ are both true, then by necessity $q$ is true too. But if you only assume that $p$ is true, you cannot conclude that $q$ is true. Indeed, in the truth table you wrote, on the second row you have that $p$ is true and $p \to q$ is false, and $q$ is false. – Taroccoesbrocco Feb 02 '21 at 07:06