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If we have $p\implies q$, then the only case the logical value of this implication is false is when $p$ is true, but $q$ false.

So suppose I have a broken soda machine - it will never give me any can of coke, no matter if I insert some coins in it or not.

Let $p$ be 'I insert a coin', and $q$ - 'I get a can of coke'.

So even though the logical value of $p \implies q$ is true (when $p$ and $q$ are false), it doesn't mean the implication itself is true, right? As I said, $p \implies q$ has a logical value $1$, but implication is true when it matches the truth table of implication. And in this case, it won't, because $p \implies q$ is false for true $p$ (the machine doesn't work).

That's why I think it's not right to say the implication is true based on only one row of the truth table. Does it make sense?

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Your distinction between "true" and "logical value 1" is not one that formal logic generally observes. Here "1" and "true" are synonyms for the same concept.

The meaning of the $\Rightarrow$ connective is what its truth table says it is, neither more nor less -- the truth table defines the connective (in classical logic). Fancy words such as "implication" or "if ... then" are just mnemonics to help you remember what the truth table is, and what the connective is good for -- but when there's a conflict between your intuitive understanding of those words and the truth table, the truth table wins over the words.

The important thing to realize is that $\Rightarrow$ is designed to be used together with a $\forall$. If you try to understand its naked truth table it doesn't seem very motivated -- certainly it can't express any notions of cause and effect, because the truth values of $p$ and $q$ just are what they are in any given world. As long as we're only looking at one possible state of the world, there's not much intuitive meaning in asking "what if $p$ held?" because that implies a wish to consider a world where the truth value of $p$ were different.

The device of standard formal logic that allows us to speak about different worlds is quantifiers. What we want to say is something like

In every possible world where I put in a coin, the machine will spit out a soda.

(though that is a little simplified -- we want to consider a "possible world" to be one where I made a different decision about my coins, not to be one where the machine had inexplicably stopped working even though it does work now. But let's sweep that problem aside for now).

This is the same as saying

In every possible world period, it is true that either I don't put in a coin, or I get a soda.

which logically becomes, using the truth table

For all worlds $x$, the proposition (In world $x$ I put in a coin) $\Rightarrow$ (In world $x$ I get a soda) is true.

Since there's a quantification going on, the truth value of the whole thing is not spoiled by the fact that there are some possible worlds with a broken machine where the $\Rightarrow$ evaluates to true. What interests us is just whether the $\Rightarrow$ evaluates to true every time or not every time. As long as we're in the "not every time" context, the machine is broken, and that conclusion is not affected by the "spurious" local instances of $\Rightarrow$ evaluating to true in particular worlds.

The construction that models (more or less) our intuition about cause and effect (or "if ... then") is not really $\Rightarrow$, but the combination of $\forall\cdots\Rightarrow$.

Unfortunately in the usual style of mathematical prose it is often considered acceptable to leave the quantification implicit, but logically it is there nevertheless. (And to add insult to injury, many systems of formal logic will implicitly treat formulas with free variables as universally quantified too, so even there you get to be sloppy and not call attention to the fact that there's quantification going on.)


Note also that this is the case even in propositional logic where there are no explicit quantifiers at all. To claim that $P\to Q$ is logically valid is to say that in all valuations where $P$ is true, $Q$ will also be true -- there's a quantification built into the meta-logical concept of "logically valid".

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  • $\begingroup$ I just wanted to clarify when a statement $p$ implies $q$ is true - the implication is false if there can be true $p$ and false $q$, even though we can have $p$ and $q$ both false at the same time (in which case the truth value of implication is $1$). So there's a difference between implication that is true in general (it's the case when we can't have $p=1$ and $q=0$), but it can be true or false with respect to specific logical values of $p$ and $q$. If I understand correctly, this is basicall what you've said. $\endgroup$ – user216094 Dec 2 '15 at 16:40
  • $\begingroup$ @user216094: Yes, I think that sounds accurate. $\endgroup$ – Henning Makholm Dec 2 '15 at 17:34
  • $\begingroup$ In other words, no matter what are logical values of $p$ and $q$, the statement $p \implies q$ tells us the possible combinations of truth values of $p$ and $q$ (all are allowed, except true $p$ and false $q$). We can assign any value truth value to $p$ and $q$ as long as there exists a "true" entry in the truth table of implication. I think it's the most compact explanation I can come up with - is it correct? $\endgroup$ – user216094 Dec 28 '15 at 18:37
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Well, yes, you can not say that any formula $\varphi$ (or more explicit $p\rightarrow q$) is allways true just based on one row of a truth table. You have to check all the rows. However If $p$ is evaluated to $True$ and $q$ is evaluated to $True$ then $p\to q$ is evaluated to $True$, thus $p\to q$ is true in this instance (also called valuation).

You have to separate between a formula being true for a valuation and allways true i.e. a tautology (or logical truth if you want to go to predicate logic).

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Implication is just like any other logical connective, and in fact, $a \to b$ can also be written using negation and disjunction as $\neg a \lor b$. Just as with logical disjunction and conjunction, it is very possible to know the value of the entire operation by observing just one operand - if $a = F$ we know $a \to b$ for all $b$, just as we'd know $a = T$ would result in $(a \lor b)$ being true for all $a$.

In your example, the implication is true if and only if the machine gives a drink (at least) when a coin is inserted. It can also spew out drinks freely at other times too ($p = F,\ q = T$) and the implication would remain true. If the machine is broken in a manner that $p = T,\ q = F$, then the implication is false, just like $(p \land q)$ would be false for the same values.

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You have to distinguish between cases where the truth values for $p$ and/or $q$ are given and cases when they're not. If the truth values for $p$ and $q$ are given then only that row of the truth table is relevant. If on the other hand they are given completely or partially all the relevant rows are relevant.

For example if we know $q$ to be true you have to consider both rows when $q$ is true - for which $p\rightarrow q$ happens to be true for all of them.

For your example one assumes that you don't insert a coin and therefore it's known that $p$ is false (and $q$ too) so you only need to consider those rows for the truth table.

$$\begin{matrix} p & q & p\rightarrow q\\ \hline f & f & T \\ f & T & T \\ T & f & f \\ T & T & T \\ \end{matrix}$$

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The fact some specific values satisfy the formula doesn't mean the formula is true in general. "It is noon now and it rains" is true right now and right here, but at other place or in other time this will appear false.

Your implication will turn out true if you check it keeps satisfied by any possible combination of its components values.

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When is implication true?

$p\implies q$ is true when both $p$ and $q$ are true, or when $p$ is false. Assuming your problem is with the latter.

Theorem:

$\neg p \implies [p \implies q]$

Proof:

Suppose $\neg p$.

Suppose $p$.

Suppose to the contrary that $\neg q$.

We have the contradiction $p\land \neg p.$

Therefore $\neg\neg q$ by contradiction.

Remove $\neg\neg$ to obtain $q$.

Therefore $p\implies q$.

Therefore, as required, $\neg p \implies [p \implies q]$.

Not sure why this is thought to not hold in natural language, but it is widely believed. Anything whether true or false will follow from a falsehood. What is wrong with: If pigs could fly, I would be the King of France. (Assuming pigs can't actually fly. See If Pigs Could Fly at my math blog.)

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