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The following is the truth table for implication $P \implies Q$.

$$ \begin{array}{|c|c|c|} \hline P & Q & P \implies Q \\ \hline T&T&T \\ T&F&F \\ F&T&T \\ F&F&T \\ \hline \end{array} $$

For $P = T$, the truth values for implication makes intuitive sense.

However, I, like many, don't intuitively understand the truth table for $P=F$.

There are many questions and answers on this site, and discussions in other places (blogs, videos, more formal courses) - but none have convinced me, and judging by the discussions there, many others also remain unconvinced.

So after some reading I think I found a rationale which I can be happy with, and I wanted to check its validity.

Comment

My reading has led me to understand that implication can mean different things - logical entailment, causal relationship, conceptual subset, material implication, conditional - all terms I am not perfectly clear on.

However, that reading has also led me to understand that as far as truth-values are concerned, this implication $\implies$ should not be seen as a causal relationship, but purely a truth-value relationship by definition, one which can usefully be interpreted as a conditional (but not causal) relationship, "if $P$ is true, then $Q$ is true".

In many real-world scenarios, there may be a causal link with $P$ and $Q$, but that is coincidental, and not strictly what this implication operator means. For example, if $P$ is "n is divisible by 6", and $Q$ is "n is divisible by 3", then we can prove the implication by means of the properties of $P$ which cause $Q$ to be true, and not by enumerating the truth table for all $n \in \mathbb{N}$, which is impossible for an infinite set anyway.

Alternatively, in a finite scenario which has been completely enumerated, such as the following, we can say "it rains" implies "eat chocolate" even though there is no causal link. The coincidence always happens.

$$ \begin{array}{|c|c|} \hline rains & chocolate \\ \hline T&T \\ F&T \\ F&F \\ \hline \end{array} $$

Question 1 - Is this comment right?

Analysis

To explain why the truth table us defined as above for $P=F$ we start with a statement we know to be self-evidently true (question 2 - an axiom?):

$$ (P \land Q) \implies P $$

That is, if $P$ and $Q$ are both true, then we can say that $P$ is true. If this wasn't axiomatically true, then we have bigger problems.

So if we agree that $ (P \land Q) \implies P $, then we can use the truth table for $(P \land Q)$ to derive the truth table for $P\implies Q$.

$$ \begin{array}{|c|c|c|c|c|} \hline row & P & Q & P \land Q & (P \land Q) \implies P \\ \hline (a) & T&T&T & T \\ (b) & T&F&F & T \\ (c) & F&T&F & T \\ (d) & F&F&F &T \\ \hline \end{array} $$

The column for $(P \land Q) \implies Q$ must always be true, because that is what we have agreed is axiomatically true.

From this we can work backwards to derive the truth table for $A \implies B$

$$ \begin{array}{|c |c|c|c|} \hline row & A & B & A \implies B \\ \hline (i) & T&T&T \\ (ii) & T&F&F \\ (iii) & F&T&T \\ (iv) & F&F&T \\ \hline \end{array} $$

Let's detail how:

  • row (i) is derived from reading row (a) that $P \land Q = T$ and $P=T$, giving $T$.
  • row (iii) is derived from reading row (b) that $P \land Q = F$ and $P=T$, giving $T$.
  • row (iv) is derived from reading row (c) or (d) that $P \land Q = F$ and $P=F$, giving $T$.

This doesn't give is row (ii) for $A = T$ and $B = F$. Here we say that we define implication to be false in this scenario because that is how we want to define it. If we defined it to be true, then there would be no meaning or use for the implication operator.

Question 3 - Is this a valid derivation of the truth table for implication?

Penelope
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    Have you searched the site for previous questions on this topic? I'd estimate that we get a question about the truth table for implication at least once a month. – Alex Kruckman Aug 10 '23 at 01:58
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    See if this helps. – David Aug 10 '23 at 02:06
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    Q1: Yes. Q2: I've never seen that statement presented as an axiom, I'd imagine it's usually presented as a tautology or Identity. Q3: Your approach is a bit misguided,p → q is a convenient shorthand for the statement: ¬ p ∨ q Once we have defined ¬ and ∨, then the truth table for → follows from that. – Michael Carey Aug 10 '23 at 02:14
  • Instead of defining (P∧Q)⟹P as a tautology in order to derive A⟹B's truth table, why not just treat the former as motivation for the definition of A⟹B (either via that first truth table or as ¬A ∨ B)? P.S. In your middle section, I don't think 'coincidence' is the correct word. – ryang Aug 10 '23 at 02:21
  • I give the exact same argument here: https://math.stackexchange.com/questions/70736/in-classical-logic-why-is-p-rightarrow-q-true-if-p-is-false-and-q-is-tr/3467187#3467187. – Bram28 Aug 10 '23 at 05:43
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    $\lnot P\lor P$ is always true. Therefore, either $\lnot P$ or $P$ is true and when $P$ is true then $Q$ is true. Therefore, we get that either $\lnot P$ is true or $Q$ is true which is $\lnot P\lor Q$. – John Douma Aug 10 '23 at 05:54
  • See https://math.stackexchange.com/q/48161/442 – GEdgar Aug 10 '23 at 07:49
  • hi @JohnDouma I like this. You start from an even more basic self-evident truth (axiom?). I'm thinking through what it means for the starting tautology if when $P$ is false, we incorrectly defined $P \implies Q$ to be false. – Penelope Aug 10 '23 at 08:17
  • thanks @MichaelCarey for answering the specific questions, especially Q1. And yes, "tautology" is a better description for Q2. – Penelope Aug 10 '23 at 08:18
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    @Penelope If $P$ being false implies $Q$ is false then, by the contrapositive, we get $Q\implies P$. Intuitively, we are saying that if $P$ is not true then $Q$ cannot be. Can you see how that makes the truth of $P$ depend upon the truth of $Q$? – John Douma Aug 10 '23 at 09:05
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    Bram28's linked answer is doing just what I suggested above (approaching it from the opposite direction as your approach): using the requirement for (P∧Q)⟹P to be a tautology to motivate the definition of A⟹B; in other words, defining A⟹B in a way so that (P∧Q)⟹P turns out to be a tautology. – ryang Aug 10 '23 at 16:39
  • I haven't read all the answers and comments, just want to add that an important property of $\Rightarrow$, i.m.o., is that for $P,Q$ with $P\Rightarrow Q$, $Q\Rightarrow P$ we have $P=Q$ (also written as $P\Leftrightarrow Q$). Take this together with your axiom $(P\land Q)\Rightarrow P$ and you get all four cases. – fweth Aug 11 '23 at 06:26
  • Also pretty sure somebody already pointed out, but you might be interested in Heyting algebras, where the implication is kind of a first class citizen. For Boolean algebras, my point of view is that you just want to construct all operations corresponding to a function $f:2^n\to 2$ with $2:={0,1}$. If you have $\neg,\land$ you can define $P\Rightarrow Q$ as $P\lor\neg Q$ but there are other sets of basic constructors which include $\Rightarrow$, e.g. $\land, \Rightarrow, 0$, because you can define $\neg P$ as $P\Rightarrow 0$. – fweth Aug 11 '23 at 06:34

1 Answers1

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Textbooks vary in the degree to which they try to motivate the truth-functional conditional (as opposed to just presenting it in a "take it or leave it" spirit!).

For one of the more extended discussions you could take a look at Chs 18 and 19 of An Introduction to Formal Logic by, erm, P_t_r Sm_th.

A PDF of the second edition (originally CUP) can be freely downloaded from https://www.logicmatters.net/ifl and those chapters can, I think, easily be read standalone: they don't depend on specific details of what's gone before.

Peter Smith
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