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It's well known that vacuous truths are a concept, i.e. an implication being true even if the premise is false.

What would be the problem with simply redefining this to be evaluated to false? Would we still be able to make systems work with this definition or would it lead to a problem somewhere? Why must it be the case that false -> false is true and false -> true is true?

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    $\begingroup$ Then implication just becomes and. This is easily seen by writing out a truth table. $\endgroup$ – Riley Mar 12 '18 at 2:50
  • $\begingroup$ Unless we made false->false true $\endgroup$ – user539262 Mar 12 '18 at 3:07
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    $\begingroup$ But then it would become a biconditional. $\endgroup$ – Bram28 Mar 12 '18 at 3:13
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    $\begingroup$ It is not a matter of "an implication being true even if the premise is false". The implication is assumed to be true. However, if the antecedent is false, then the truth value of the consequent is not constrained. $\endgroup$ – Rodrigo de Azevedo Mar 12 '18 at 12:16
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    $\begingroup$ Vacuous truth normally means ∀x ⋲ {} x ⊨ Q; that is all statements about all members of the empty set are true. $\endgroup$ – Joshua Mar 12 '18 at 15:50
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Notice that 3=5 is false. but if 3=5 we can prove 8=8 which is true.

$$ 3=5$$

therefore $$ 5=3$$

Add both sides, $$8=8$$

We can also prove that $$ 8=10$$ which is false.

$$ 3=5$$

Add $5$ to both sides, we get $$8=10$$

The point is that if we assume a false assumption, then we can claim whatever we like.

That means " False $\implies$ False " is true.

And " False $\implies$ True " is true.

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    $\begingroup$ This really made it click for me. From a false premise we can arrive at both true and false statements alike. But from a true premise we can't arrive at false conclusions, only true ones. Is this right? $\endgroup$ – user539262 Mar 12 '18 at 3:23
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    $\begingroup$ Right., I am glad that a simple example made the point through. $\endgroup$ – Mohammad Riazi-Kermani Mar 12 '18 at 3:28
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    $\begingroup$ But how would you go from $3=5$ to 'snow is white' or to 'snow is purple'? You have shown a specific false claim and a specific true claim that we can infer from a false claim, but I don't see how that proves the general: 'if we make a false assumption, then we can claim anything we like'. Or, as the OP put it: if we use different variables for $P$ and $Q$, then maybe we cannot go from $P$ to $Q$, even if $P$ is false. $\endgroup$ – Bram28 Mar 12 '18 at 3:44
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    $\begingroup$ @Bram28 Once, Berttand Russell said in a lecture that, if 1+1 = 1, then everything can be proven true. Someone challenged him, “Then prove you’re the Pope.” Russell said, “One and one is one. I am one. The Pope is one. Therefore, I and the Pope am one.” $\endgroup$ – Davislor Mar 12 '18 at 5:46
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    $\begingroup$ @Stilez This isn't what Gödel's theorem says. Starting from true premises always leads to true conclusions. Gödel says that formal systems powerful enough to express arithmetic cannot prove all true statements. In particular, they cannot prove their own consistency. You do, I guess, have to allow for the possibility (however faint) that Peano arithmetic is inconsistent. But then, we are still not deriving false from truth: we have been all the time been deriving false (and truth) from falsity. $\endgroup$ – Steven Gubkin Mar 12 '18 at 14:41
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Clearly we want $P\rightarrow P$ to be true, wouldn't you agree?

I mean, if i say:

If Pat is a bachelor, then Pat is a bachelor

do you really dispute the truth of that claim, or claim that it depends on whether or not Pat really is a bachelor? The whole point of conditionals is that we can say 'if', and thereby imagine a situation where something would be the case, whether it is actually the case or not. And guess what: if Pat would be a bachelor, then Pat would be a bachelor, even if Pat is not actually a bachelor.

So, if $P$ is false, it better be the case that $false \rightarrow false = true$, for otherwise $P \rightarrow P$ would be false, which is just weird.

Of course, we also want $true \rightarrow true = true$ by this same argument, for otherwise again we would have $P \rightarrow P$ being false.

As far as $false \rightarrow true$ is concerned: given that we have that $true \rightarrow true =true$, $false \rightarrow false$, and ( I think you would certainly agree) $true \rightarrow false = false$, we better set $false \rightarrow true =true$, because otherwise the $\rightarrow$ would become commutative, i.e. We would have that $P \rightarrow Q$ is equivalent to $Q \rightarrow P$ ... which is highly undesired, since conditionals have a 'direction' to them that cannot be reversed automatically. Indeed, while I think you would agree with the truth of:

'if Pat is a bachelor, then Pat is male'

I doubt you would agree with:

'if Pat is male, then Pat is a bachelor'

EDIT

Re-reading your question, and considering some of the ensuing discussions and comments, I wonder if the following might help:

Suppose that we know some statement $P$ is false, i.e. We know that:

$1. \neg P \quad Given$

Then we can show that $P$ implies any $Q$, given the standard definition of logical implication:

$2. P \quad Assumption$

$3. P \lor Q \quad \lor \ Intro \ 2$

$4. Q \quad Disjunctive \ Syllogism \ 1,3$

And, using our typical rule for $\rightarrow \ Intro$, we can then also get:

$5. P \rightarrow Q \quad \rightarrow \ Intro \ 2-5$

And this of course works whether $Q$ is true or false.

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  • $\begingroup$ "Clearly we want $P\rightarrow P$ to be true" I'm not sure of this? Why would we necessarily want it to be true? Why couldn't $P\rightarrow P$ be false when $P$ is false? $\endgroup$ – user539262 Mar 12 '18 at 3:02
  • $\begingroup$ "We would have that P→Q is equivalent to Q→P ... which is highly undesired." Why undesired? $\endgroup$ – user539262 Mar 12 '18 at 3:05
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    $\begingroup$ @user539262 The $\rightarrow$ is supposed to capture conditionals, i.e. 'If ... then ...' statements. Frankly, I can't think of a more clear example of an 'if ... then ..' statement being true than if the 'if' and 'then' parts are the very same. $\endgroup$ – Bram28 Mar 12 '18 at 3:07
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    $\begingroup$ @user539262 Also, conditionals have an inherent 'direction' about them. It makes sense that 'if Pat is a bachelor, then Pat is male', but not that 'If Pat is male, then Pat is a bachelor' $\endgroup$ – Bram28 Mar 12 '18 at 3:09
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    $\begingroup$ @user539262 True. In fact, here is an example: 'If Pat lives in Paris, then Pat lives in Germany'. Assuming Pat lives in London, we have a false anteceddent, a false consequent, and (I agree) a false conditional. hey, no one said that the $\rightarrow$ perfectly captures the English conditional, and in fact examples like this show that the eNglish conditional simply isn't truth-functional. But, for formal logic purposes, we want our operator to be truth-functional. And, out of all the options, the one we chose works the best, for reasons I indicated, and further reasons yet. $\endgroup$ – Bram28 Mar 12 '18 at 3:33
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I don't have a lot to say on this, but I used to be very annoyed by the concept of vacuous truth and only these two observartions soothed my ailment.

1.) One clearly wants $A\land B\implies A$, and this wouldn't be true without $false \implies true$ being true.

2.) $false \implies true $ is exactly the same statement as "the empty set is contained in every other set" which to me is intuitive.

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  • $\begingroup$ Why would $A \land B \implies A$ be something one clearly wants? $\endgroup$ – user539262 Mar 12 '18 at 3:04
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    $\begingroup$ @user539262 Well, if A is true, and B is also true, then A is true. $\endgroup$ – JKEG Mar 12 '18 at 3:08
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    $\begingroup$ I don't understand how the two statements are exactly the same in 2. Could you please explain? $\endgroup$ – Eric Duminil Mar 12 '18 at 8:17
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    $\begingroup$ @Eric I think the argument goes something like this: let $X$ and $Y$ be sets, and $z$ be an instance of the type of thing contained in these sets. Then $X\subset Y$ is defined as ($\forall z: z\in X\implies z\in Y$). If $X\subset Y$ is to be true when $X = \emptyset$, then because $z\in\emptyset$ is always false, we need $\text{false}\implies z\in Y$ to be true for all $z$'s, both those in $Y$ and those not in $Y$, which in turn requires both $\text{false}\implies\text{true}$ and $\text{false}\implies\text{false}$ to be true. $\endgroup$ – David Z Mar 13 '18 at 1:45
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Right now, we like making statements like

If $a$ and $b$ are both even, then $a+b$ is even.

Or, we could write this mathematically as $$ a \equiv b \equiv 0 \pmod{2} \implies a+b \equiv 0 \pmod{2}. $$ If we were to redefine $\implies$ to disallow vacuously true implications, this would no longer be a true statement, because of cases such as "$1$ and $3$ are both odd, but $1+3$ is even". But we still want to talk about such statements, so we'd probably just end up saying longer sentences such as

The statement "$a$ and $b$ are both even" implies in the sense that allows vacuously true implications the statement "$a+b$ is even".

This is a very useful relationship to talk about, so you'd just condemn us to longer phrasing for no good reason. Meanwhile, we already have conjunctions such as "and" and "iff" to describe cases where both statements must be true, or where both statements must have the same truth value.

Mathematical terminology is driven by utility. If it were useful to have "if" mean the thing you want it to mean, we'd do it. But it's useful to have if-then statements be true in vacuous cases, just like it's useful to (to give some other examples) have $0$ be an even number and have $1$ be neither prime nor composite. So we go with the meaning that makes our lives easier.

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It's well known that vacuous truths are a concept, i.e. an implication being true even if the premise is false.

What would be the problem with simply redefining this to be evaluated to false?

One answer to the question is that we would no longer be able to know that PQ from the fact that we know ¬PQ. We would also not be able to know that PQ from ¬(P ∧ ¬Q). And we would no longer know from ¬Q → ¬P that PQ.

So, for example if I knew that either a given number was not divisible by 3, or that it was even, I would not be able to deduce that if the number was divisible by 3 it would be even.

Similarly, if I knew that a given number was not both divisible by 3 and not even, I would similarly not be able to infer that if the number was divisible by 3 it would be even.

Lastly, if I knew that if a given number was not even it was also not divisible by 3, then I would not be able to know that if it was divisible by three, it was even.

In each of the cases above, if If P, Q was false when P was false, the conclusion could be false because the number in question might be divisible by 3.


Note

If we made PQ true when P was false then → would just be ∧. We can't replace material implication (MI) with ∧ because, MI is what it is by definition. I therefore take this question to mean what would happen if we took 'If P, Q' to be false when P is false.

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The assumption here is that we are working with a bivalent logic in which all statements are true or false. They are never both true and false, nor are they neither true or false.

Now if this is your universe, let's see...

  1. Everything is either true or false.

  2. Now assume false (is true).

  3. We already know that true is true, so if false is ALSO true, ...

  4. Then everything is true!

This answer doesn't say how you can derive any true statement from a false statement. It just says that if all false things were true, then all things are true (assuming you want everything to be false or true).

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    $\begingroup$ The principle of explosion is commonly assumed even in non-classical logics that admit richer semantics than truth tables. $\endgroup$ – Derek Elkins Mar 13 '18 at 3:43

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