# What's the difference between material implication and logical implication?

When I read the definitions of material and logical implications, they seem to me pretty much equivalent. Could someone give me an example illustrating the difference?

(BTW, I have no problem with the equivalence between $\lnot p \vee q$ and $p \to q$, aka "if $p$ then $q$". My confusion is with the idea that there are two different forms of implication, material and logical.)

Thanks!

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They are indeed identical. The term "material implication" is supposed to distinguish implication, in the logical sense, from the informal notion of implication, which carries some sense of connection. –  André Nicolas Oct 1 '11 at 2:38

There is one level at which they can be distinguished. The following definitions are relatively common.

• Material implication is a binary connective that can be used to create new sentences; so $\phi \to \psi$ is a compound sentence using the material implication symbol $\to$. Alternatively, in some contexts, material implication is the truth function of this connective.

• Logical implication is a relation between two sentences $\phi$ and $\psi$, which says that any model that makes $\phi$ true also makes $\psi$ true. This can be written as $\phi \models \psi$, or sometimes, confusingly, as $\phi \Rightarrow \psi$, although some people use $\Rightarrow$ for material implication.

In this distinction, material implication is a symbol at the object level, while logical implication is a relation at the meta level. In other words, material implication is a function of the truth value of two sentences in one fixed model, but logical implication is not directly about the truth values of sentences in a particular model, it is about the relation between the truth values of the sentences when all models are considered.

There is a close relationship between the two notions in first-order logic. It is somewhat immediate from the definitions that if $\phi \to \psi$ holds in every model then $\phi \models \psi$, and conversely if $\phi \models \psi$ then $\phi \to \psi$ is true in every model. This relationship becomes more fuzzy when we begin to look at other logics, and in particular it can be quite fuzzy when philosophers talk about material conditionals and logical implication independent of any formal system.

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@AsafKaragila: it's not clear to me how the statement after "in particular" follows from the theorem. What is $T$ in this particular case? –  kjo Oct 1 '11 at 12:30
@Asaf: That complicates things, because then you have to talk about provability. Also, not every logical system satisfies the deduction theorem. (Also, you stated the converse of the actual deduction theorem, which says that if $\alpha \vdash \beta$ then $\vdash \alpha \to \beta$; the converse you stated is essentially modus ponens.) I thought about it and decided against it. –  Carl Mummert Oct 1 '11 at 12:31
@Carl: I see. Thanks for the correction anyway. –  Asaf Karagila Oct 1 '11 at 13:47
Isn't there another form of logical "implication", since ϕ⊨ψ means we have ψ as a semantic consequence of ϕ, so ϕ implies ψ in a semantic sense, while ϕ|-ψ means we have ψ as a syntactic consequence of ϕ, so ϕ implies ψ in a syntactic sense? If not, why is "ϕ|-ψ" not also an implication? –  Doug Spoonwood Oct 23 '11 at 0:16
What is "an implication" in general? At least by convention, we don't usually use the term "implication" for the $\vdash$ relation. –  Carl Mummert Oct 23 '11 at 21:54