Can anyone exhibit a mathematical sentence in which a conditional (not necessarily the main connective) has to be STRICTLY understood as a MATERIAL one, and would become false if the material conditional was understood as logical implication instead?

Some context:

In logic, strictly speaking, material implication (' → ') has to be carefully distinguished from logical implication (' ⇒ '). However, I have noticed that in mathematics books, the distinction is not emphasized, as if, in that field, all implications are logical implications. Is it actually the case? (Reference at Archive.org: On this distinction and on the symbols I use , Seymour Lipschutz, Schaum's Outline of Set Theory , ch. 14 " Algebra Of Propositions".)

To illustrate the difference between material and logical implication, consider the sets A={ x | x is a mathematician → x is a musician } and B={ x| x is a mathematician ⇒ x is a musician }. A is simply the set of people who (contingently) happen not to be both mathematician and non-musician, since its conditional is a material one. However, B is the set of people such that for each member, it is or would have been logically impossible for them to be mathematician without being musician; depending on one's opinion concerning the relationship between mathematics and fine arts, one will probably tend to answer either that B is either the universal set (a mathematician is necessarily a musician) or the empty set.

I think that substituting ' → ' for ' ⇒ ' cannot lead to important problems, since, if A logically implies B, then A should also materially imply B ("A ⇒ B" meaning that (A → B) is true in all possible cases, all possible "interpretations"). Here I'm asking the reverse question: is it always correct to substitue ' ⇒ ' for ' → ' in mathematics, in other words, is it correct to use always " ⇒ " in mathematics?

My question is not on symbols.

  • $\begingroup$ Logical implication usually is material implication. $\endgroup$ Mar 21, 2019 at 17:24
  • $\begingroup$ Here is a helpful reference on typesetting mathematical symbols. $\endgroup$
    – Théophile
    Mar 21, 2019 at 17:35
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    $\begingroup$ In mainstream mathematics, there is only one type of implication. I've never heard of the distinction you mention, but it sounds like the one type is material implication, because if it is true that Paris is the capital of France, then the implication is true. $\endgroup$ Mar 21, 2019 at 17:37
  • $\begingroup$ @MattSamuel If you have never heard of it, with what authority are you saying that "there is only one type of implication"? Actually, you are wrong that the concept of material implication doesn't exist in mathematics. $\endgroup$
    – user647486
    Mar 21, 2019 at 17:48
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    $\begingroup$ Could shed some light on the issue math.stackexchange.com/questions/68932/… $\endgroup$
    – chhro
    Mar 21, 2019 at 18:12

3 Answers 3


The symbols you want are $\to$ (\to) for material implication and $\implies$ (\implies) for logical implication. Insofar as mainstream mathematics distinguishes them, $p\implies q$ means that $p\to q$ is (a) true in all models of a theory of interest (however, in that context we'd usually write $\models$ (\models) instead of $\implies$ to make it clear) or (b) a tautology. And in modal logic, we can rewrite $p\implies q$ as $\Box(p\to q)$ (note the use of \Box). But in practice, $\implies$ is often used in proofs to indicate an inference from what was already known.

  • $\begingroup$ Thanks for the clarification concerning symbols and underlying concepts. How would you symbollically translate statement such that : " if 4² is even then 4 is even ". With the (\Box) or without it? $\endgroup$
    – user655689
    Mar 21, 2019 at 22:26
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    $\begingroup$ @RayLittleRock Always make your statement only as strong as you intend. The statement $2|4^2\to 2|4$ will be implied whatever symbols you use, as all alternatives to $\to$ are at least as strong. If you want to make the further statement that all models yield the above, change $\to$ to $\models$; if you deem it "necessary" (whatever you take that to mean metaphysically), by all means wrap the statement in $\Box()$; if you dare claim definitions have sufficed to make it a tautology, use $\implies$. $\endgroup$
    – J.G.
    Mar 21, 2019 at 22:29

Consider the statement

"If x is even, then x is divisible by $2$"

In 'math world' we regard this statement as true.

But this is not a logical truth. That is, logically one is allowed to interpret 'even', 'divisible by' and '$2$' in a way that would make the statement false.

So, the statement is a mathematical truth, but not a logical truth. More to the point: the 'if' part does not logically imply the 'then part. Indeed, if we were to symbolize it, we should be using the material implication, and not the logical implication.

Of course, if we are given the (normal!) definitions of 'even', 'divisible by' and '$2$', then we can logically infer the truth of the statement above as a whole. That is, the statement as a whole is logically implied by the relevant definitions.

Also, if we fill in a specific value for $x$, say $4$, then the statement becomes:

"If $4$ is even, then $4$ is divisible by $2$"

And now, given the standard definitions/axioms (let's refer to that as a set of statements $A$), we have that $A$ together with "$4$ is even" logically implies that "$4$ is divisible by $2$" ... but we still don't have that "$4$ is even" by itself logically implies that "$4$ is divisible by $2$"

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    $\begingroup$ @RayLittleRock Right. Or to be exact: when a mathematician makes an 'if ... then .. ' claim, it is rarely a claim of logical implication. $\endgroup$
    – Bram28
    Mar 21, 2019 at 18:44
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    $\begingroup$ @ryang Yeah, well noted! The OP’s usage of logical implication in their example was rather weird, so I decided to use my own example. $\endgroup$
    – Bram28
    Feb 24, 2023 at 22:12
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    $\begingroup$ @PW_246 It would be something that I can logically derive from some set of axioms that I postulate about some domain, but that does not make it a logical truth ... it would merely be a truth-in-that-theory. A logical truth is something I can derive from no axioms at all, i.e. something that would be true-in-any-theory $\endgroup$
    – Bram28
    Jun 2, 2023 at 14:52
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    $\begingroup$ @PW_246 Logic is all about patterns. "If all gluckeroos are smibbel, then all gluckeroos are smibbel" logically makes sense. From a logic point of view, this is of the form "If P then P". On the other hand, from a purely logical point of view, "If n is even, then is divisible by 2" is of the form "If P then Q", so it is not a logical truth. You need to provide meaning and content and background assumptions and axioms (i.e. math) to see that "If n is even, then is divisible by 2" is true. $\endgroup$
    – Bram28
    Jun 2, 2023 at 18:02
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    $\begingroup$ @PW_246 Of course it can be shown to be true. But not by logic alone. Like you say, you need axioms that define the notions of ‘even’, ‘2’, and ‘divisible’. As soon as you do that, you go beyond pure logic. That’s what makes this a mathematical truth, not a logical truth. $\endgroup$
    – Bram28
    Jun 3, 2023 at 2:13

In mathematics based on classical logic, there appears to be no difference between material and logical implication.

Consider, for example, the implication: If it is raining ($R$), then it is cloudy ($C$).

$$R \implies C$$

This implication does not mean that rain causes cloudiness, or that cloudiness causes rain. It means only that, at the moment, it is not both raining and not cloudy.

$$\neg [R \land \neg C]$$

This is often used in one form or another as The Definition of $\implies$ in introductory textbooks, but it can also be derived from other widely accepted properties of implication, conjunction and negation in classical logic:

  • Introducing and Eliminating $\land$
  • Eliminating $\neg\neg$
  • Conditional proof
  • Proof by contradiction
  • Detachment (Modus Ponens)

See my formal proof.

So, in classical logic anyway, the above "definition" would seem to apply to every implication.

  • $\begingroup$ Neat proof. Thanks! How do you understand this equivalence of ( A-->B) and ~ (A& ~B) $\endgroup$
    – user655689
    Mar 22, 2019 at 17:37
  • $\begingroup$ @RayLittleRock I see it as applying in mathematics and in natural language as well, provided you are talking about a pair of unambiguous true-or-false propositions at a given moment in time. In mathematics, of course, there is no notion of the passage of time, no future or past, essentially only the present -- that which IS true. $\endgroup$ Mar 22, 2019 at 17:58

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