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So far I've understood that not every "If" in natural English can be symbolized by the material implication ⊃ for truth-functionally

But then, I find in several books an argument "for ⊃" stating that:

"If A then B" is true iff A ⊃ B is true.

Then it develops proving that statement. I agree that they are logically equivalent but I do not see how this might be an argument for ⊃ being a good representative of the "if" in English.

As I understand it, in order to develop the proof of the statement you need to assume "If A then B" has the nature of the material implication. So I think of it as an attempt to define a word with the same word.

Why would it be a good argument?

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    $\begingroup$ Please provide an example of such an argument. $\endgroup$ – Y. Forman Jan 8 '18 at 12:44
  • $\begingroup$ @Y.Forman There's a couple such arguments below! -- fairly standard textbook stuff, I think. $\endgroup$ – Peter Smith Jan 8 '18 at 17:22
  • $\begingroup$ @PeterSmith Yup, I figured the question was referring to something like that (or other variants I've seen; e.g., assuming "if A then B" must have a simple truth table and then showing that it could not be anything but that of material implication), but it's been a little while since I've seen this stuff, so I wanted to hesitate to comment on something nonspecific. Your answer does a good job of doing just that, though, so +1 to you. $\endgroup$ – Y. Forman Jan 8 '18 at 17:30
  • $\begingroup$ Do you mind providing an example of your first statement? "So far I've understood that not every "If" in natural English can be symbolized by the material implication ⊃ for truth-functionally " What are some examples of the mismatch of material implication in logic and natural language? $\endgroup$ – Pinocchio Jan 21 '18 at 16:40
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The usual arguments for assigning simple indicative conditionals the same truth-conditional content as the material conditional are variants of the following.

Argument 1:

(i) Suppose if A then C. So we either have not-A, or we have A and hence C. So if A then C implies either not-A or C.

(ii) Conversely, suppose we are given either not-A or C. Then if not the first, then the second. So we can infer if A then C.

Hence if A then C is equivalent to the truth-functional not-A or C.

Argument 2

(i) The claim if A then C rules out having A true and C false. So if A then C implies it isn’t the case that both A and not-C.

(ii) Conversely, suppose we are given that it isn’t the case that both A and not-C. Then we can infer that if A is actually true we can’t have not-C as well: in other words if A then C.

Hence if A then C is equivalent to the truth-functional not(A and not-C).

But each of not-A or C and not(A and not-C) is (uncontroversially) equivalent to the material conditional we symbolize A $\supset$ C. Hence if A then C is true if and only if the corresponding A $\supset$ C is true.


Now, I'm not claiming that these arguments are irresistible. But we can make three points

  1. We do have some substantial arguments here: these are not e.g. bogus "attempt(s) to define a word with the same word".
  2. If you are going to reject the identification of the basic, truth-relevant content of 'if A then C' with the material conditional, you need to explain where Arguments (1) and (2) go wrong....
  3. ... and you need to explain too why Arguments (1) and (2) have looked as compelling as they do to many logicians.

But equally, of course, if you do accept these Arguments (1) and (2), then you have other tasks, basically to explain away why the identication of 'if' and '$\supset$' leads to such counter-intuitive results in certain cases!

The literature on both sides of the debate is enormous, and inconclusive! My view, for what it is worth, is that Frege's attitude in his Begriffsschrift, at the birth of modern logic, is the right one. Treat (what we now call) the material conditional as a workable substitute for the ordinary conditional for mathematical purposes, and don't worry too much about its exact relation to messy ordinary usage.

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  • $\begingroup$ Thanks for the reply! What I am understanding is that arguments (1) and (2) show that the material implication is truth-functional, since you manage to prove logical equivalence to another proven truth-functional notion (-A V B). But then at the end, that argument is no trying to prove that '⊃'is an accurate representation of the natural English "If". We are assuming it is the best available representation of the natural counterpart and we are showing that it is truth-functional thus consistent with our formal system. Am I I right? PS: I love your book and your website! $\endgroup$ – César D. Vázquez Jan 9 '18 at 13:37
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    $\begingroup$ @CésarDalíVázquezNavarro "[A]rguments (1) and (2) show that the material implication is truth-functional". No, material implication is truth-functional by definition -- no argument needed! "The argument is no trying to prove that is an accurate representation of the natural English "If"" No, the arguments DO aim to show that the truth-relevant content of "If"" (in indicative conditoinals) is just $\supset$. [We could still argue that "if" has additional non-truth-relevant content -- compare "but" versus "and", arguably same truth-relevant content but different overall meanings.] $\endgroup$ – Peter Smith Jan 9 '18 at 13:44
  • $\begingroup$ I understand that there are a lot of arguments for ⊃ being the best way to symbolize "If" in natural English. I am not saying that is not the case, but what I do not understand is how can arguments 1 and 2 be proof of that. I was expecting a more linguistic/meaning/philosophical argument. Thanks again $\endgroup$ – César D. Vázquez Jan 9 '18 at 13:52
  • $\begingroup$ The claim (from defenders of those sorts of arguments) is that e.g. (1) shows that just appealing to the ordinary inferential behaviour of "if" plus excluded middle shows that "if A then C" is interderivable with "Either not-A or C". So if you think that meaning is tied to role in inference, this IS a philosophical argument appealing to meaning! $\endgroup$ – Peter Smith Jan 9 '18 at 13:57
  • $\begingroup$ I will read more about it and come back with more specific questions. Thanks for your time! $\endgroup$ – César D. Vázquez Jan 9 '18 at 14:09
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My personal but ever more strongly held view is that relating formal logic to natural language is something an introductory book on logic (and most books on logic) should not do. Many books do do this, but, in my opinion, this just creates unnecessary confusion, a misrepresentation of how logic is used by most mathematicians, and in some cases is just meaningless. I do agree that linguistics is an application area of logic, but the picture of this presented in most (all?) introductory logic texts that take this approach is utterly facile.

To start, if this is the "definition" of $A \supset B$ you were given, then it is completely inadequate. Even when we are applying logic to lingistics, the goal is to understand natural language utterances by translating them to logic formulas. This is of no use if the logical formulas are "defined" by natural language utterances as you point out. When we're mathematicians and logicians and not linguists, then "If $A$, then $B$" is simply not unambiguous nor something that can be mathematically manipulated.

Now, what you can do is talk about how to read logical formulas. You can say $A\supset B$ is read as "if $A$, then $B$". This is a unidirectional relationship and generally not meant as a definition, but merely as a communication aid. For example, I can say $x+y$ can be read as "add $x$ to $y$" but 1) this clearly doesn't explain anything about addition, and 2) this doesn't mean "add Tom to the invitation" means $\text{Tom}+\text{the invitation}$.

There are two ways of defining the logical connectives. One approach is syntactic where we give rules and axioms that show how a connective can be used. For example, one way of defining $\supset$ is via the following two rules: $$\cfrac{\Gamma, A\vdash B}{\Gamma\vdash A\supset B}\supset\!\!I\qquad\cfrac{\Gamma\vdash A \supset B \quad \Gamma\vdash A}{\Gamma\vdash B}\supset\!\!E$$ What's actually happening here is these two rules form part of a definition of a binary relation we've notated as $\vdash$. The second rule, $\supset\!\!E$, states (in the informal metalogic) that if $\Gamma \vdash A\supset B$ and $\Gamma \vdash A$ hold, then $\Gamma\vdash B$ holds. The collection of all such rules gives an inductive definition of the $\vdash$ relation. Informally, $\Gamma\vdash A$ means "$A$ is provable given assumptions $\Gamma$". This is, again, just a way to read the notation or at best a guide to the desired interpretation. The definition is the collection of rules.

The second approach of defining logical connectives is semantic. Here we systematically map logical formulas into mathematical objects. The most well-known such semantics is one for classical propositional logic where formulas are mapped into Boolean functions which are usually presented as "truth tables". Here the interpretation of a formula $A\supset B$, written $[\![A\supset B]\!]$ must be of the form $[\![\supset]\!]([\![A]\!],[\![B]\!])$ and the interpretation of $\supset$ is $$[\![\supset]\!](x,y)=\begin{cases}0, &\text{if }x=1\text{ and }y=0\\1, &\text{otherwise}\end{cases}$$ Many people take issue with this definition when they first see it. Part of this is due to the use of loaded terms like "true" and "false". Another part is that this semantics is only adequate for classical propositional logic. More complicated semantics are needed for more complicated logics like classical first-order logic. In general, different logics will require different notions of semantics, e.g. truth tables are not suitable for intuitionistic propositional logic. What this definition does accomplish is reducing the meaning of logical formulas to a clear and unambiguous notion that can be manipulated mathematically. For those who do want to translate natural language to logical formulas, they will need to disambiguate and spell out nuances in the translation, but they will at least have a solid target whose consequences can be calculated to cross-check.

In some cases these two approaches are connected by soundness and completeness (meta)theorems. These theorems together imply that a given syntactic notion of proof/derivability coincides with a given semantic notion of validity.

So, to answer your question, there is no good argument because 1) this is not how implication is defined, 2) matching the natural language "if" is not the goal, and 3) $\supset$ is indeed not a good representation for the natural language "if". The issue is natural language is messy, ambiguous, and inconsistent. We clearly don't want those properties to apply to our formalization, so the primary goal of mathematical clarity overrides any secondary goal of corresponding to natural language. This pushes the complexity of mapping a natural language statement into a logical formula to the translation process where it won't impact consumers of logic that aren't interested in linguistics. Even for linguists, it makes sense to have a precise language where different interpretation of superficially similar or even the same statement look different.

In practice, classical propositional logic is way too simple to capture, in any way, the meaning of many natural language statements even ignoring nuance. Different and more powerful logics are typically used. It is a completely reasonable goal to want to build a logic that more closely captures natural language utterances, but this is a goal at the intersection of linguistics and logic, not the goal of logic as a whole. There are plenty of reasons why the definitions given above are useful mathematically that are completely unrelated to natural language.

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  • $\begingroup$ I wish I could upvote this more than once. $\endgroup$ – Henning Makholm Jan 9 '18 at 22:31
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    $\begingroup$ Part of the problem is that glossing $A\to B$ as "if $A$ then $B$" really only makes any natural-language sense if $A\to B$ is being said in a context where it's either implicitly or explicitly quantified over something -- whether that "something" is "all times" or "all possible worlds" or a variable that appears in $A$ and $B$. And this is specifically not the case in the pure propositional calculus where most textbooks attempt to introduce the connective. $\endgroup$ – Henning Makholm Jan 9 '18 at 22:40
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Let $A$ and $B$ be true-or-false, logical propositions. (Edit: Assuming the Law of the Excluded Middle) For both material implication and if-then constructs in natural language, we know that:

  1. The Deduction Rule: If we assume $A$ is true and can subsequently determine, without making any other assumptions, that $B$ is also true, then we can infer that $A$ implies $B$ ($A \implies B$ ). If we can also determine that $B$ is false ($\neg B$), then we can also infer that $A$ is false ($\neg A$).
  2. The Detachment Rule: If both $A \implies B$ and $A$ are true, then $B$ will also be true.

It then seems inevitable to me that we would also then have the propositions:

  1. $A \land B \implies [A \implies B]$
  2. $A \land \neg B \implies \neg[A \implies B]$
  3. $\neg A \implies [A\implies B]$
  4. $[A\implies B] \iff \neg[A \land \neg B]$

(Edit: (1-3) are the basis for the usual truth table for implication. (4) is often given as The Definition of material implication.)

The proofs of each of these propositions is a trivial application of the rules of natural deduction (see If Pigs Could Fly). And these propositions pretty much define both material implication as well as if-then constructs in natural language for any true-or-false propositions $A$ and $B$.

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  • $\begingroup$ "It then seems inevitable to me". Well, it perhaps shouldn't seem just inevitable, if that means indisputable by any reasonable logician! -- , because both your (initial) 1 and 2 are intuitionistically acceptable, while your 4 [now corrected for the obvious typo] isn't. $\endgroup$ – Peter Smith Jan 9 '18 at 16:26
  • $\begingroup$ Thanks for the edit. If you require the proofs -- quite short for formal proofs -- see "If Pigs Could Fly" at my math blog dcproof.wordpress.com $\endgroup$ – Dan Christensen Jan 9 '18 at 16:50
  • $\begingroup$ The proofs are standard textbook stuff. I'm just pointing out that getting (4) from (1) and (2) requires excluded middle, so is only as "inevitable" as excluded middle. And some would certainly dispute that that is inevitable! $\endgroup$ – Peter Smith Jan 9 '18 at 16:54
  • $\begingroup$ Recall, I am talking about "true-or-false propositions." (My first and last lines,) $\endgroup$ – Dan Christensen Jan 9 '18 at 16:58
  • $\begingroup$ Ah -- OK, fair enough, if that indeed means "assuming excluded middle" (as opposed to "propositions with determinate contents that can be true or false") $\endgroup$ – Peter Smith Jan 9 '18 at 17:03

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