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RESOLVED: I fundamentally just didn't understand implication. Thank you to everyone who helped!

I was going off of my linguistic understanding of "implies", which isn't the same. For example, the sentence "this [true thing] implies that [another thing] could be false" makes grammatical sense, and logically fits many scenarios, but it doesn't align with the truth table of logical implication.

For me personally, the interpretation that matches the actual truth table for implication is as follows: read X -> Y backwards, and phrase "implies" as "facilitates".

i.e. "Y facilitates X".

Y being true facilitates X to be true, but doesn't make any guarantees:

  • false -> true = makes perfect sense (Y happened and X didn't; this satisfies the definition of implication; the expression is true);
  • true -> true = makes perfect sense (Y happened and X did; this satisfies the definition of implication; the expression is true)).

Y being false offers no facilitation, so this prevents X being true:

  • false -> false = makes perfect sense (Y didn't happen so X can't; this satisfies the definition of implication; the expression is true));
  • true -> false = is illogical (Y didn't happen so X can't, but, if it still does, this violates the definition of implication, and so the original expression is false)).

I'm confused by an example of basic logic symbols on Wikipedia. The example under the arrow symbol "implies" is this:

If P and Q are logical predicates, P ⇒ Q means that if P is true, then Q is also true. Thus, P ⇒ Q is logically equivalent with Q ∨ ¬ P.

I get the general relationship being implied (in situations where P is true, Q is also true), but I'm confused at the rephrasing to Q ∨ ¬ P. As I read it, I end up with the following possibilities (a basic OR logic table):

  • false OR not false [true] = true
  • true OR not false [true] = true
  • true OR not true [false] = true
  • false OR not true [false] = false

In that case, Q can be true without P being true, and P can be true without Q being true. That would mean the original example isn't the total working of the relationship, just "incidentally, the two can be true simultaneously". Also, then, that the only thing that matters it that one is true, and thus the order of P and Q, and the presence or absence of ¬, therefore doesn't seem that important. This is all uncomfortably broad, so I've probably missed something.

Or is this just specific to this example? I.e., here, is P ⇒ Q just indicating a subset of potentialities in a wider, unexplained scenario, in such a way that (just sometimes) Q ⇒ P and P ⇔ Q can also be true, and that there is no obvious relationship (at least not at this level of understanding of the problem)?

That would seem very cosmic for a basic explanation. If that's meant to be a bit of a trick example, I'll take it as extra education to not just assume things. But, as this isn't a test - just free education - I'm much more inclined to believe I've grossly misunderstood.

Please can someone help explain this to me?

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  • $\begingroup$ Your logic table is correct, but you may be helped by writing it as $\neq P \vee Q$, so that the propositions are in the same order as $P \implies Q$. What you're finding out is that the statement $P \implies Q$ is false if and only if you try to make something true imply something false. You cannot reach a false conclusion from a true premise. $\endgroup$
    – prets
    Dec 5, 2021 at 14:50
  • $\begingroup$ Anything else is fair game---true implies true (that's hopefully not surprising); false implies true (take e.g. the statement $1 = 0$ and multiply both sides by $0$, then it implies the true statement $0 = 0$); false implies false (add, say, $1$ to each side of the previous example). $\endgroup$
    – prets
    Dec 5, 2021 at 14:51
  • $\begingroup$ Note also that it isn't the same as the logic table for $P \vee Q$ since the false entry for it will be in a different row. $\endgroup$
    – prets
    Dec 5, 2021 at 14:53
  • $\begingroup$ "[...] the order of $P$ and $Q$, and the presence or absence of $\neg$, therefore doesn't seem that important" --- They're hugely important. Unlike $\vee$, $\implies$ isn't commutative. $P \implies Q$ and $Q \implies P$ are not the same things. $\endgroup$
    – prets
    Dec 5, 2021 at 14:55
  • $\begingroup$ (My first comment has a \neq instead of a \neg, sorry for the typo. It should say $\neg P \vee Q$.) $\endgroup$
    – prets
    Dec 5, 2021 at 14:56

1 Answer 1

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(Making big tables in comments is tedious...)

The logic table for $P \implies Q$ is not the same as that of $P \vee Q$ (they are not equivalent); they have the same number of trues and falses, sure, but not in the same places: $$ \begin{array}{c c | c} P & Q & P \implies Q \\ \hline \text{true} & \text{true} & \text{true} \\ \text{true} & \text{false} & \color{red}{\text{false}} \\ \text{false} & \text{true} & \text{true} \\ \text{false} & \text{false} & \text{true} \\ \end{array} \qquad \begin{array}{c c | c} P & Q & P \vee Q \\ \hline \text{true} & \text{true} & \text{true} \\ \text{true} & \text{false} & \text{true} \\ \text{false} & \text{true} & \text{true} \\ \text{false} & \text{false} & \color{red}{\text{false}} \\ \end{array}$$

The demonstrates that the presence or absence of $\neg$ in $\neg P \vee Q$ certainly matters, and so does the order of $P$ and $Q$:

$$ \begin{array}{c c | c} P & Q & Q \implies P \\ \hline \text{true} & \text{true} & \text{true} \\ \text{true} & \text{false} & \text{true} \\ \text{false} & \text{true} & \color{red}{\text{false}} \\ \text{false} & \text{false} & \text{true} \\ \end{array} $$

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