# Using "implies" to refer to material conditional

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 $|f(x)-f(c)|<\epsilon$.

• 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. Commented Apr 7, 2023 at 20:35

• $$\huge\rightarrow$$ (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.

• 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? Commented Jan 15, 2021 at 5:22
• So if you’re proving an “if... then” or “implies” statement in math, you’re proving a statement about propositions/meta-proposition, correct? Commented Jan 15, 2021 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). Commented Jan 15, 2021 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? Commented Jan 16, 2021 at 4:56
• @ryang Ah ... that's why this old post popped up! :P OK, yes, we're on the same page :) Commented May 19, 2022 at 13:11

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 $\neg p$.

• 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. Commented Nov 17, 2016 at 8:12
• @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 $q$. Commented Nov 17, 2016 at 8:17
• @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). Commented Nov 17, 2016 at 8:38
• @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. Commented Nov 17, 2016 at 8:55
• @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. Commented Nov 17, 2016 at 17:49