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Is it acceptable to translate the binary connective "$\let\ f\rightarrow$" into English with "implies"? I'm unsure because "implies" for me immediately brings to mind logical implication, but I've seen some places use it for the material conditional (including wikipedia, in the opening sentence of this article).

For example, does mathematical convention, in principle, permit the following formulation of the standard definition for functional continuity?

$f$ is continuous at $c$ if for any $\epsilon > 0 :$ there exists $\delta > 0:$ for any $x$ in Domain[$f$]$:$ $|x-c|<\delta$ implies $|f(x)-f(c)|<\epsilon$.

ryang
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CuriousKid7
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    The duplicate target discusses the difference between the material conditional and logical implication (for the uninitiated, this means implication regardlesss of model, and does NOT refer to the implication that we use in mathematics). On the other hand, this question is asking about the difference between the material conditional and mathematical implication. Thus, the duplicate target DOES NOT at all address this question. – ryang Apr 07 '23 at 20:35

2 Answers2

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  • $\huge\rightarrow$ (the material conditional) is a logical connective that operates on its antecedent $P$ and its consequent $Q$ to form the truth function $$P\rightarrow Q,$$ which is false precisely when $P$ is true but $Q$ false.

  • $\huge\Rightarrow$ (implication) similarly operates on its antecedent $P$ and its consequent $Q;$ here, the result $$P\Rightarrow Q$$ lives in a specific (usually implicitly understood) interpretation/context and signifies that $P\rightarrow Q$ is true in it. Some ways to read $P\Rightarrow Q:$

    • If $P$ is true, then $Q$ is true.
    • $P$ being true is a sufficient condition for $Q$ to be true.
    • $P$ being true implies that $Q$ is true.
    • $P$ is true only if $Q$ is true.
    • $Q$ being true is a necessary condition for $P$ to be true.

    When $P\not\Rightarrow Q,$ then it must be that $P$ is true yet $Q$ false.

In the given formulation $$|x-c|<\delta\;\; \textbf{implies} \;\;|f(x)-f(c)|<\varepsilon,$$ “implies” is not the material conditional $\large\rightarrow\normalsize$ per se, but rather mathematical implication $\large\Rightarrow\normalsize;$ it analytically (from mathematical axioms and a given context) asserts that its right side can be derived from its left.

P.S. I distinguish between implication $\,\large\Rightarrow\,$ and logical implication $\,\large\vDash\,,$ which is often used to mean first-order implication, i.e., that $P\rightarrow Q$ is true regardless of interpretation.

P.P.S. Symbolic logic is an area rife with conflicting notation, terminology and even notions; my understanding is eclectically evolving.

ryang
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  • In the Wikipedia article, it states that a valid interpretation of the material conditional is “if p is true, then q is also true,” which you say is a way to read “p implies q.” Is the Wikipedia article incorrect or am I misinterpreting what you’re saying? – Adam French Jan 15 '21 at 05:22
  • So if you’re proving an “if... then” or “implies” statement in math, you’re proving a statement about propositions/meta-proposition, correct? – Adam French Jan 15 '21 at 15:46
  • @Adam: Yes and no. You can think of it as a meta-statement, but you can also think of it as equivalent to the material implication, which is now an internal statement. The key point is that the two turn out to be equivalent (in the standard scenarios anyway, I'm not sure what happens in non-classical logics). – Asaf Karagila Jan 15 '21 at 21:26
  • @AsafKaragila Can you think of them as equivalent because showing the consequent follows from the antecedent(assuming the antecedent is true) guarantees the conditional is true? – Adam French Jan 16 '21 at 04:56
  • @Adam: They are equivalent because of soundness, completeness, and truth tables. – Asaf Karagila Jan 16 '21 at 11:50
  • @AsafKaragila okay, thank you! – Adam French Jan 16 '21 at 15:40
  • For future readers, this is a complementary explanation of a similar viewpoint as this post. – ryang Jul 21 '21 at 07:17
  • @ryang I would say that the ‘implies’ used here is one of mathematical implication, by which I mean logical implication under the assumption of various mathematical axioms. So it is not a pure logical implication, and so I would not use $\implies$ if I had to formalize this. Indeed, if I had to formalize this, I would use the material $\to$… in combination with the formalizations of the aforementioned axioms you get what you want if you ever wanted to use this for doing proofs. – Bram28 May 19 '22 at 12:53
  • @Bram28 Yes, precisely; your comment is consistent with my Answer including its first postscript; in fact, I had just today hinted the same here and here. – ryang May 19 '22 at 13:09
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    @ryang Ah ... that's why this old post popped up! :P OK, yes, we're on the same page :) – Bram28 May 19 '22 at 13:11
  • @ryang Hey, can you please reconsider the closing of this question: https://math.stackexchange.com/questions/4878212/when-can-compound-biconditional-statements-be-true-when-the-values-of-the-indiv Yes,the OP used poor language, but the OP did provide a good example to elaborate the question, and I am not sure what else the OP could have done: it is a very straightforward question about a rather unintuitive result. Indeed, it is a common question that my students run into, and the answer reveals an all important distinction between the way we use $\leftrightarrow$ vs $\Leftrightarrow$. – Bram28 Mar 11 '24 at 12:09
  • @ryang Well, his title is pretty clear (which is really the only thing that's clear from his post :P ), and I have had this question from students many times before. :) – Bram28 Mar 11 '24 at 17:25
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Yes.

By the way, logical implication is material conditional. In logic only the forms of the arguments matter in order to deduce from something.

When you see $$ |x-c|<\delta\text{ implies }|f(x)-f(x)|>\epsilon $$ written in a proof, it is certainly an English version of the formal statement $$ |x-c|<\delta\to|f(x)-f(x)|>\epsilon. $$

Sometimes authors say at the beginning of their book that the proofs will be given in an informal manner. Informal means that English language will be used for better readability. In principle, those informal proofs could be made formal in, say, first-order logic.

Note. The logical connective $\to$ really contains what we mean by "implies". Indeed, $p\to q$ does what it is supposed to do: it permits us to infer $q$ from $p$ but nothing from $\neg p$.

Guest
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  • Logical implication is not the same as the material conditional. The material conditional of the form "if p then q" is a logical connective that can be replaced in a proof by "not-p or q." It is just a disjunction. But logical implication refers to entailment, which is a relation between two sentences. Also, what you're referring to in your note is a rule of inference, not a logical connective. – CuriousKid7 Nov 17 '16 at 08:12
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    @CuriousKid7 If you say so. But if that is what you call logical implication, then it is not a well chosen name. In the context of mathematical logic, the implication $p\to q$ absolutely discards any a priori relation between the two propositions $p$ and $q$. – Guest Nov 17 '16 at 08:17
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    @CuriousKid7 My note says, indeed, that the logical connective $\to$ gives, in particular, the modus ponens rule of inference, hence justifying in that sense the fact that it contains what we mean by "implies" in ordinary language. But as you know that is not enough. For example, you could infer $q$ from $p$ and $p\wedge q$, but also from $\neg p$ and $p\wedge q$. This is not in accordance with the intuitive meaning of "implies", so it would be counter-intuitive to call the logical connective $p\wedge q$ an implication or a conditional (since there is no condition anymore). – Guest Nov 17 '16 at 08:38
  • Thank you the oringal answer about the conventional use of "implies." But I think we're talking past each other now. Our ordinary use of the conditional is not really at issue here. Nor is what the material conditional captures about what me mean by "implies." My point was that, formally, the material conditional is just a particular disjunction formed from the antecedent and consequent. Also, the material conditional cannot "give" a rule of inference like modus ponens--unless you mean that modus ponens justifies the inference "q" from "if p then q" and "p." – CuriousKid7 Nov 17 '16 at 08:46
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    @CuriousKid7 I feel you're just playing with words here. Also, you say that your issue is not about what the material conditional captures about what we mean by implies. What's the first paragraph of your post about, then? In that case I fear I don't understand what you're asking or why you're asking, since you seem so knowledgeable. – Guest Nov 17 '16 at 08:55
  • I was worried about ambiguity in the way I described, but I was just asking about whether or not mathematicians find it acceptable as a matter of convention. It was just a superficial question, really--which you definitely settled. – CuriousKid7 Nov 17 '16 at 09:45
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    @CuriousKid7 Yes, it is acceptable as a matter of convention, since it is written in the literature that $p\to q$ represents the statement "$p$ implies $q$" or "if $p$ then $q$" or "$p$ only if $q$", etc. – Guest Nov 17 '16 at 17:49