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Imagine the following conditional:

If washing machines are humans, washing machines are quadrupeds.

It seems to me that the truth value of the conditional as a whole is ambiguous. Since its antecedent is false, logic tells us that the conditional is (vacuously) true. But in fact, the conditional as a whole does seem false: if we grant that washing machines are humans, then washing machines are clearly bipeds.

But how can the same conditional be both vacuously true and (at least intuitively) false?

Xander Henderson
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Sokito
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    If $P$ is false then $P$ implies everything, just as a matter of truth tables. in your case, there is no instance of a human washing machine that is not a quadruped, so the claim is (vacuously) true. – lulu Sep 02 '23 at 11:52
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    How we deal with such statements in daily life does not matter. In math , it is clearly defined. We would even consider some $A\implies B$-statements not as meaningful , even if $A$ and $B$ are both true , but if there is no link bewteen $A$ and $B$. But again , this is irrelevant for the logical value of the statement. – Peter Sep 02 '23 at 12:16
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    The point (computationally speaking) of vacuously true statements is that you'll never be able to get a washing machine that is a human, so it doesn't matter what you put afterwards. – Julián Sep 02 '23 at 12:30
  • Both your conditional statement and "If washing machines are humans, washing machines are bipeds" are vacuously true, if you know what "washing machines" are. But only the second statement can be derived from the statement "humans are bipeds"* without defining "washing machines".

    *(Sorry, I am aware that not all humans are bipeds)

     – peterwhy Sep 02 '23 at 13:36
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    The notion of a conditional statement in logic, has very little to so with conditionals, used in natural language. Your intution here, is grounded in natural language- which is misguiding you. In common speech, implication presumes two events are related in some way. Whereas, in logic, We can form a implication between any two statements - regardless of if they are related. If 1 = 0, then I am a donkey, is nonsense in common parlance, but a true statement in logic. – Michael Carey Sep 02 '23 at 15:50
  • You would have better luck posting this in a philosophy forum, as you're really asking about counterfactuals in natural language. – blargoner Sep 02 '23 at 15:57
  • This is called a counterfactual conditional, as opposed to the usual material conditional found in logic. There is a nice theory of these by David Lewis. For a counterfactual conditional, the falsity of the antecedent does not imply the conditional. – Joshua Tilley Sep 02 '23 at 22:37
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    @JoshuaTilley Actually, it isn't a counterfactual according to Lewis. The counterfactual version would be "If washing machines were humans, washing machines would be quadrupeds". Lewis had to have three different theories of conditionals, one for 'indicatives' like OP's (material implication); one for counterfactuals (possible worlds); one for indicatives with "adverbs of quantification" (something approaching normal humans' interpretations of indicative conditionals - that gave rise to Kratzer's theories) – Araucaria - him Sep 02 '23 at 23:54
  • @blargoner No this is not a counterfactual. A counterfactual would be "If washing machines were humans, washing machines would be quadrupeds". Your comment is a good sign that the material implication theory of conditionals [which many people assume is the only theory of conditionals used within a logic system, as if there was only one unified system of logic] does not apply to the logic of natural languages. It's precisely because washing machines aren't humans that OP's sentence is true according to a truth-functional analysis. – Araucaria - him Sep 03 '23 at 00:16
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    @Araucaria Yes I'm well aware of all that, thank you. That's why I didn't say what the OP wrote was itself a counterfactual, but what OP is really [meaning to] ask about are counterfactuals. – blargoner Sep 03 '23 at 00:21
  • @blargoner No, I don't think so at all. Counterfactuals aren't vacuously true. The OP is just being pulled in two directions because their natural language logic (which conforms more closely to the mathematical philosophy of Gibbard or Stalnaker, and includes a principle of conditional non contraditiction) tells them that it must be false. In the same way that when you aren't being primed by an MI theory of conditionals you would disagree with a conditional that said that if the number in my hand is odd it's divisible by 2, even if you did not know whether the number was odd or not. – Araucaria - him Sep 03 '23 at 00:27
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    @Araucaria You're completely misinterpreting my comments but I don't care to engage further. Take care. – blargoner Sep 03 '23 at 00:28
  • @blargoner I'm keen to understand if I've misunderstood. How is the OP asking about counterfactuals? I'm missing something somewhere. – Araucaria - him Sep 03 '23 at 00:30
  • @Peter But there are several different logics that have been devised to deal with conditionals over the years, and there are different. mathematical ways of dealing with them. – Araucaria - him Sep 03 '23 at 00:50
  • In classical logic, no statement can be both true and false. – Mauro ALLEGRANZA Sep 04 '23 at 06:10
  • To everyone involved in the edit war here: stop. When the original poster rolls back an edit, you believe this is in error, raise a flag. If someone repeatedly edits your question in a way you don't like, raise a flag. Back-and-forth rollbacks are unproductive. – Xander Henderson Sep 14 '23 at 14:41
  • short answer: no – RyRy the Fly Guy Sep 14 '23 at 16:17

6 Answers6

7

But in fact, the conditional as a whole does seem false: if we grant that washing machines are humans, then washing machines are clearly bipeds.

It is indeed true that if machines are humans, then washing machines are bipeds. But from this you can not infer that the other conditional is false. They are both true at the same time, precisely because they are only vacuously true. Two vacuously true statements with the same false antecedent are not contradictory. See also Why is it that the statement "All goblins are yellow" does not contradict the statement "All goblins are pink?" Neither of the two statements is false formally logically speaking, even though this may seem unintuitive.

Natalie Clarius
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Intuition is a tricky thing and does not always play well with mathematics. Plenty of mathematical results are non-intuitive.

As lulu says in a comment: If $P$ is false then $P$ implies everything; that's just how it is in mathematics, most of the time (1).

Once I was in the office a little after the end of day and a manager said: "why is no one still here?". I replied: "they have just gone for a dinner break; any time that I have been here at midnight, it was very busy". He clearly did not want to accuse me of lying but, also, he clearly did not believe me.

Here is a simple example of day to day usage which does not match mathematical usage. If I say "do you want tea or coffee?" then I probably do not expect the answer "both"; in common usage, "or" is usually exclusive. In the world of mathematics, "both" would be acceptable; "or" is inclusive unless specified otherwise.

A less simple example: intuitively there are more positive integers than just even positive integers yet, in mathematics, we say that there are equally many.

(1) There is no standards body for mathematics so there are no absolute definitions and rules. Some are followed by most people most of time time. Others vary a lot from author to author. In this case, alternative systems of logic are studied.

badjohn
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The word "if" has two different meanings. The meaning that's used in mathematical writing is different from the meaning that's used in everyday English.

In mathematical writing, the phrase "if X, then Y" is defined as meaning "either not-X, or Y." Whenever this definition conflicts with the everyday meaning of the word "if," we disregard the everyday meaning and use this definition instead.

So if we come across the sentence "If washing machines are humans, then washing machines are quadrupeds" in a mathematical text, then this sentence is defined as meaning "Either washing machines are not humans, or washing machines are quadrupeds," and this sentence is clearly true.

It's true that according to the everyday, intuitive meaning of the word "if," the sentence "If washing machines are humans, then washing machines are quadrupeds" is probably false*. But that fact is simply irrelevant, because in mathematical writing, we just don't use the everyday, intuitive meaning of the word "if."

(*Actually, I might argue that we can count the number of feet a washing machine stands on, and it's clearly four, and therefore, if washing machines are humans, then they must be humans who stand on their hands and feet, and are therefore quadrupeds. But I digress, I digress!)

Tanner Swett
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  • Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed. – Xander Henderson Sep 12 '23 at 22:31
1

  1. This conditional—in its entirety—is not false:

    If washing machines are humans, then washing machines are quadrupeds.

  2. On the other hand, given that

    • washing machines are humans,

    it is indeed false that

    • washing machines are quadrupeds.

Thus, under the assumption of the antecedent (pretending that it is actually true), we can assert that the consequent—not that the entire conditional—is false. Here, the consequent's falsity is in the context of a subroutine, so to speak.

The word ‘implication’ informally has two conflicting meanings in the English language: sometimes it means the conclusion/consequent, other times it means the entire conditional. When the antecedent and consequent are both false, these two meanings should be carefully distinguished, because in this case the consequent (the ‘implication’) and the entire conditional (also the ‘implication’) have different truth values.

In any case, these two statements don't actually contradict each other:

If washing machines are humans, then washing machines are quadrupeds.

if we grant that washing machines are humans, then washing machines are clearly bipeds.

The sentences $$H{\implies}Q$$ and $$H{\implies}\lnot Q$$ are, by definition, simultaneously (vacuously) true.

Appending my comment from Araucaria's answer

The statements

  • given event $A,$ the (conditional) probability of event $B$ is $0.6$

and

  • if event $A$ happens, then the probability that event $B$ happens is $0.6$

have different meanings: the former says that $$P(B|A)=0.6$$ whereas the latter says that $$\text{A happens}\implies P(B)=0.6.$$

When $A$ denotes the empty set and $B$ denotes the sample space, the first statement is false while the second statement is vacuously true.

ryang
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Washing machines are not humans, so how can you know whether they would be biped or quadruped if they were? In a universe where washing machines are humans, perhaps humans are quadrupeds!

You say:

"if we grant that washing machines are humans, then washing machines are clearly bipeds."

but this is only a guess, not a certitude. It could only be a certitude if washing machines were humans, which they aren't.

This is why we say that if A then B is true if A is false: because if A then B can only be falsified if A is true.

Stef
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  • I'm pretty sure the OP made up a crazy conditional statement on purpose to see if their claim is vacuously true or not. – Accelerator Sep 15 '23 at 10:36
  • @Accelerator Indeed; but I fail to see the logical link between your comment and the downvote. – Stef Feb 14 '24 at 13:57
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The fact that there is a mathematical theory (one amongst many) of how to represent natural language conditional statements, and it is well known by those maths students who have never studied natural language conditionals, this does not mean that that theory works as a model of the logic of natural language conditionals! It doesn't. For example, any mathematical theory of probability will disintegrate on contact with a material implication theory of natural language conditionals as proposed by Bertrand Russell.

For example, consider a situation where I throw a fair dice two times. What is the probability that, if I throw a six the first time, I will throw a six the second time?

The natural and correct answer to this question is 1/6, of course. However, iff the material implication theory of conditionals meted out to maths students is correct, then that answer is completely and utterly bananas. The answer according to the material implication theory of conditionals is 31/36.

Mathematicians will only every give you an answer to a maths question according to the material implication theory of conditionals if they have been primed to do so first. They need this priming so that they can relegate their ability to understand speech to the status of a false theory given to toddlers, and then supplant this understanding of natural language conditionals with a theory that mathematicians love to talk about but never actually use in real life or real maths.

There are other mathematical models of natural language conditionals. See for example Ernest Adams, Dorothy Eddington, Allan Gibbard. They work much better. According to theories like these conditionals have no truth values, but have assertibility conditions, where the assertibility of a conditional If P, Q is equivalent to the probability of Q given P. This kind of theory of conditionals doesn't implode on impact with either common sense or any mathematical theory of conditional probability.

On this kind of understanding of conditionals, the Original Poster's conditional might be "vacuously true" according to the material implication theory of conditionals, but in actual fact is a sentence that's incapable of having truth conditions. It would be unassertible because the conditional probability of Q being true given P in this instance is basically zero.

There are many other mathematical theories of natural language conditionals that would say that this conditional was false (see for example this famous paper by Robert Stalker: Probability and Conditionals 1970 or the paper Indicative conditionals 1975.

Araucaria - him
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    I disagree that "given event A, the (conditional) probability of event B is $0.6$" means "if event A happens, then the probability that event B happens is $0.6$". The former says that $$P(B|A)=0.6,$$ whereas the latter says that $$\text{A happens}\implies P(B)=0.6.$$ – ryang Sep 03 '23 at 01:56
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    Considering the question was asked on a math forum, rather than a philosophy or linguist forum, I think it's fair to assume that the OP is working inside the standard first order logic framework, that every student taking Discrete Math works in. And going into other theories may just be confusing for the OP. This is like, someone insisting that 2 + 2 = 5, to a middle schooler, if we choose to work in a particular modulus. I don't know, how appropraite the main takeway is, or how this alleviates the OP's confusion. – Michael Carey Sep 03 '23 at 01:57
  • @ryang Perfectly naturally, you've let the "probability" skip over the "if A" part of the conditional. The original was "What's the probability of if A, B", not "If A, what's the probability of B"! I'd say those are the same, but others wouldn't. In any case, in both your examples, would this not be calculated as P(A^B) /P(B)? And aren't they therefore the same? – Araucaria - him Sep 03 '23 at 02:30
  • @MichaelCarey I'll cogitate. I can't tell if the thrust of your comment is correct - i.e. whether my post is counterproductive for OP (I don't currently think it is). – Araucaria - him Sep 03 '23 at 02:31
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    I don't know if my comment is correct either!! In the event that they ( or any reader is), I hope my comment brings them some clarity/direction. Regardless, I do believe your answer brings a valuable insight. – Michael Carey Sep 03 '23 at 03:07
  • "Let's throw a die and denote the event of a negative result and the event of a positive result by A and B, respectively; then the first assertion is false whereas the second assertion is true. " <-- I have no idea what you mean. For example, what do you mean by "a positive result"? – Araucaria - him Sep 07 '23 at 16:23
  • @Araucaria $\quad$ 1[rewrite]. "In both your examples, would this not be calculated as ${P(A\cap B) /P(B)}?$" $\quad$ No: to see that the two assertions in my first comment have different meanings, let $A=$ the event of a negative-numbered outcome on a die throw$={}$ and $B={1,2,3,4,5,6},$ so that the first assertion is false while the second assertion is vacuously true. – ryang Sep 10 '23 at 05:46
  • @ryang What do you mean by "the event of a negative numbered outcome on a die throw"? b) Whatever that means, how would that render the first proposition false? c) If "a negative numbered outcome" is possible, then B is not 1/6 d) where did 0.6 come from? e) you've not been faithful to the original sentence: "What is the probability that [If P, Q]" - You've transmogrified it to "If P, what's the probability that Q". I'm happy to keep discussing after that's all cleared up. – Araucaria - him Sep 10 '23 at 14:53
  • @Araucaria 1. Our die-throw experiment has six possible outcomes, none a negative number, so $A$ is an impossible event, so $P(B|A)$ does not equal $0.6.\quad$ 2. I will basically copy my above comment: the point was that a conditional probability is not quite about implications/conditionals. And it's not really meaningful to discuss the probability of "if A then B", which is not an event; but if you mean the probability that it is true, then in the context of a probability experiment, the answer can either be 0 or 1. $\quad$ I am unlikely to have more to add about this, if you don't mind. – ryang Sep 10 '23 at 15:35
  • @ryang The whole point of the truth-functionalist/Bertrand Russell/Philo of Megara account is *precisely* that the conditional can be given truth values according to the truth values of its individual clauses. And on this very basis it clearly can be given a probability. However, it's not *my* understanding of conditional probability, it's the one used by mathematicians in real life. It's also the one defended by some of the most famous philosophers of the C20th, for example Robert Stalnaker, Dorothy Edgington etc. Nothing to do with me. – Araucaria - him Sep 10 '23 at 15:38
  • @ryang Kind to let me know (I don't have anything against people downvoting posts they disagree with). The reason it relates to conditional probability is that we can give a probability to each of the four possible combinations of truth values for P and Q (T/T, T/F, F/T, F/F). So on a truth-functionalist account P(p-->q) is 1 - (the probability of the truth values of P and Q being (T/F)). You might enjoy reading these: Stalnaker 1970 Probability and Conditionals, Lewis 1976, and §5 here – Araucaria - him Sep 10 '23 at 16:19
  • @ryang And lastly (took me a while to find) Edgington 1986 Do conditionals have truth conditions? <-- mostly about conditionals and probability. – Araucaria - him Sep 10 '23 at 16:24
  • @Araucaria No, none of this, about the proposed $P(A\text{ happens}\to B\text{ happens}),$ or somesuch, and its possible interpretations, is *conditional probability*, and indeed, my initial comment was just taking issue with your answer's second last line calling either it, or the $x$ in $(A\text{ happens}\implies P(B)=x),$ "conditional probability". – ryang Sep 11 '23 at 02:22
  • @ryang When A means "x happens", it would be strange to represent a conditional as ((x happens) happens) ---> ((y happens) happens)). Anyhow, conditional probability is often defined as P(B|A), where P(B|A)=P(A^B)/(A). The penultimate sentence merely refers to the fact that P(A^B)=0 in OP's example. – Araucaria - him Sep 11 '23 at 07:20
  • @Araucaria This is pulling teeth. I've consistently been using $A$ to represent an event (a set) and $x$ to represent a number in $[0,1].$ And for the umpteenth time, whether your post's penultimate sentence is referring to $P(A∩B)$ or $P(A\text{ happens}\to B\text{ happens})$ or the $x$ in $(A\text{ happens}\implies P(B)=x),$ calling any of these 3 objects a "conditional probability" is simply incorrect. I shan't have more to add (ok, this time, really). – ryang Sep 11 '23 at 08:06
  • @ryang And from Lewis's paper (one of the most renowned papers probability and conditionals) linked-to above, entitled 'Probabilities of Conditionals and Conditional Probabilities' (p. 297): "In the case of ordinary indicative conditionals, it seems that assertability goes instead by the conditional subjective probability of the consequent, given the antecedent. We define the conditional probability function P (-/-) by a quotient of absolute probabilities, as usual: (1) P(C/A)= [df] P(CA)/P(A), if P(A) is positive" <--- that OK? – Araucaria - him Sep 11 '23 at 13:45