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I am really lost trying to understand the difference between the logical connective "implies", $\to$, and the metalogical symbol (or maybe it's also a connective?) $\vdash$. (This is all focusing on prepositional logic here).

In metalogical terms, for example with modus ponens, it is said that $P, P \to Q \vdash Q$, which means "If we have a proof of $P$, and we have a proof of $P \to Q$, then we can infer / make a proof of $Q$". But I don't understand what the difference is between that and saying something like $P \land (P \to Q) \to Q$ which is similar but uses $\to$ instead of $\vdash$.

For example the $P \to Q$, at least in my experience, means "it is possible to go from $P$ to $Q$" but I don't see how "going to $Q$" is different from "inferring $Q$." Simply telling me that one is metalogical and one is not doesn't really help me understand what's going on.

I've also been given the example of what the tortoise said to Achilles but I don't understand this, either. It sounds like the tortoise is constantly rejecting implications because "who says I have to accept conclusions just because the premises are true?" but then somehow introducing a metalogical $\vdash$ solves this? "We use the metalogical symbol $\vdash$ to basically force that stubborn tortoise to accept the conclusions and we've now circumvented the issue."

Unless I've grossly misunderstood something I just don't see why that's even a thing. Who says then I have to accept $\vdash$? Is $\vdash$ just a stronger form of $\to$, like a "sudo $\to$" or something (to borrow a Linux term), a form of implies that forces the conclusion to be accepted from the premise(s)?

What's the difference? How are they working here? Why do we need them? Are there any concrete examples showing the difference of both?

David C. Ullrich
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user525966
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  • First basic difference: we may have a logical language without the connective $\to$ (for classical logic $\land$ and $\lnot$ are enough) but the relationS of $\vdash$ (derivability in the calculus) and $\vDash$ (logical consequence) will not change. – Mauro ALLEGRANZA Sep 03 '18 at 13:55
  • But then using your connectives $\land, \lnot$ what stops us from defining $p \lor q = \lnot(\lnot p \land \lnot q)$, and $p \to q = \lnot p \lor q$ and introducing the symbols all the same? – user525966 Sep 03 '18 at 13:57
  • I don't think this is a duplicate -- I've seen that answer already and all it says is basically "this one's metalogic, this one's within logic" which does not address my confusion – user525966 Sep 03 '18 at 13:59
  • $\lnot p \lor q$ means (in the "standard" classical (i.e. truth-functional) reading of the connectives) : "either $p$ is false or $q$ is true". – Mauro ALLEGRANZA Sep 03 '18 at 14:00
  • Yes but then in mathematical proofs we normally use $\to$ to indicate moving from premises to conclusions – user525966 Sep 03 '18 at 14:00
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    See also the post Implies vs Entails vs Provable – Mauro ALLEGRANZA Sep 03 '18 at 14:02
  • I have seen that answer too, it says the same general idea: One's metalanguage, one's within the language. It doesn't really make the distinction between what it means to "conclude" something given premises, or what "provable" means in terms of being able to go from one thing to another. Like if I have a proof of $P$ and a proof of $P \to Q$ I don't automatically have a proof of $Q$, there's some intermediate step needed, but then to me this is just pushing the problem down another step. – user525966 Sep 03 '18 at 14:04
  • In a previous comment chain I was told differently: https://math.stackexchange.com/a/2903228/525966 (towards the bottom of this comment chain) – user525966 Sep 03 '18 at 14:06
  • Is an inference a sort of "blind or forced jump/movement"? As in, if I have a proof of $P$, and I also have a proof of $P \to Q$, can I now immediately say I have a proof of $Q$, or is there "more work / more steps" that need to be done to say I have proven $Q$? – user525966 Sep 03 '18 at 14:10
  • Yes; if I have a proof of $P$, and I also have a proof of $P→Q$, then - by modus ponens - I have a proof of $Q$. This means that the derivation $P, P \to Q \vdash Q$ is a one-line derivation. – Mauro ALLEGRANZA Sep 03 '18 at 14:19
  • @MauroALLEGRANZA Are operators defined in terms of the axioms and inferences we provide? For example my impression is that it is possible to construct/prove everything without even using truth tables, correct? Like if I said my language doesn't use connectives $\to, \lnot$ but two completely made up symbols, "op1" and "op2" or something, the axioms and inferences would also be in terms of those ops to define how they work? – user525966 Sep 03 '18 at 14:25
  • Does this answer your question? Implies ($\Rightarrow$) vs. Entails ($\models$) vs. Provable ($\vdash$) – user1147844 Feb 24 '23 at 08:01

3 Answers3

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First, I'm surprised that nobody has pointed out that reading $\vdash$ as "infers" is simply wrong: implies versus infers.

You might read $\vdash$ as "proves" or "entails". On the other hand, "infers" is roughly the same as "deduces". Saying $A\vdash B$ means that one can deduce $B$ from $A$; if you read $\vdash$ as "infers" you're reading $A\vdash B$ as "$A$ deduces $B$", which, regardless of whether it makes any sense, certainly does not mean the same thing.

On the difference between $\to$ and $\vdash$: $A\to B$ is just a formula in ur formal system; it does not say anything (it's not an assertion). On the other hand, $A\vdash B$ is a statement about the formulas $A$ and $B$; it says that given $A$ there is a proof of $B$ in whatever formal proof system we're taking about.

If the proof system is sound and complete then $A\vdash B$ is equivalent to "$A\to B$ is a tautology". But jumping from there to the conclusion that $A\vdash B$ is equivalent to $A\to B$ is wrong; "$A\to B$ is a tautology" is a statement about $A$ and $B$, while $A\to B$ is simply not a statement at all.

An analogy from algebra: if $x$ and $y$ are numbers then $x>y$ is a statement about $x$ and $y$, while $x-y$ is not a statement at all, it's just a number. It is true that "$x>y$ is equivalent to $x-y>0$", but if you concluded that "$x>y$ is equivalent to $x-y$" that would be clearly nonsense. Going from the true fact "$A\vdash B$ is equivalent to the statement that $A\to B$ is a tautology" to "$A\vdash B$ is equivalent to $A\to B$" is making exactly the same error

David C. Ullrich
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    "$A \to B$ is not a statement at all" What? As far as I know, it's still a statement in that the "is true" is simply omitted, i.e. $A \to B$ is the statement "$A \to B $ is true". (as per Tao's Analysis, Vol 1) – user525966 Sep 03 '18 at 15:58
  • In the present context $A\to B$ is not a statement at all. It's a good thing that you mentioned Tao, because that explains where the confusion comes from. In fact the citation from Tao is irrelevant to the present question, because it's taken from a totally different context - Tao is not doing formal logic.[...] – David C. Ullrich Sep 03 '18 at 16:18
  • When a mathematician doing math other than formal logic writes $A\to B$ they typically mean "$A$ implies $B$". They're using $\to$ as a verb; what they mean by $A\to B$ is the same as what we mean by $A\vdash B$ in formal logic. Which of course causes confusion. But in formal logic $\to$ is not a verb, and $A\to B$ is not a statement, it's just a wff. The meaning of any given symbol depends on the context... – David C. Ullrich Sep 03 '18 at 16:21
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    I admit I am really struggling to understand the difference, or why it's not considered a statement, or why the context is different, etc. – user525966 Sep 03 '18 at 16:32
  • "Why the context is different": Analysis is not formal logic. "Why it's not considered a statement": It's not considered a statement because in fact it is not a statement. This really is the important point regarding the difference between $A\to B$ and $A\vdash B$ - as long as you insist on thinking of $A\to B$ as "saying something" it will be impossible to get this straight. Read the "analogy" at the bottom of my answer! – David C. Ullrich Sep 03 '18 at 16:42
  • I did read it -- the problem is not being able to understand it or why these differences are the way they are, what suddenly changes across contexts, what these symbols are actually saying, how we're defining things in these systems, etc. In Tao's Analysis he would say $x-y$ is an "expression" whereas $x-y > 0$ would be a statement/proposition (has a true/false answer). I know you're saying "this isn't analysis" but where's the distinction? Why are the rules suddenly changing? How to keep this straight? – user525966 Sep 03 '18 at 16:47
  • "What suddenly changes across contexts": Suppose I said I was trying to understand why "Jack und Jill" makes sense in German but not in English - what "suddenly changes" across contexts? In formal logic the symbol $\to$ in $A\to B$ is not saying anything. Trying to understand why the meaning of the symbol is what it is is exactly like trying to understand why "und" in German means the same as "and" in English - there is no "why" there, symbols mean what they mean. – David C. Ullrich Sep 03 '18 at 17:02
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    Do you mean that the $\to$ is just an arbitrary symbol relating $A$ and $B$ in some way but we have no idea what until we move $A \to B$ to some logic system and assign it meaning? – user525966 Sep 03 '18 at 17:04
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    Yes, precisely! (In fact what I mean is that the $\to$ means even less than "an arbitrary symbol relating $A$ and $B$ in some way". In a careful development of mathematical logic a wff is just a sting of symbols. There's a definition that specifies what strings are or are not wffs - part of the definition is "if $A$ and $B$ are wffs then $A\to B$ is a wff". Nothing about "relating $A$ to $B$"; wffs have no "meaning" until we assign truth values.) – David C. Ullrich Sep 03 '18 at 17:16
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    So then $\vdash$ is defining what "proving" means, even when no meaning has been assigned yet, even before we get into concepts like true and false? So with modus ponens, given two wff's $A$ and $A \to B$, we have now "proven" the wff $B$, and that's all this "rule of inference" means? – user525966 Sep 03 '18 at 17:22
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    When I say that the $\to$ does not "relate $A$ and $B$": Analogously, the $+$ sign in $2+3$ does not "relate $2$ and $3$'. The $+$ sign is not a relation, it is an operator that takes two numbers and produces another number. If you want to understand the difference between $\to$ and $\vdash$ you need to understand that in exactly the same way $\to$ is not a relation, it is an operator tat takes two wffs and produces another wff, and that's all it is. Otoh $\vdash$ is a relation. – David C. Ullrich Sep 03 '18 at 17:32
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    Yes. Again, a formal proof system is just a set of rules that says that we may "derive" a wff from other wffs; it doesn't necesarily have anything to do with the "meaning" of anything. For the specific proof system you have in mind it turns out that $A\vdash B$ is equivalent to $A\models B$, in other words $v(B)=T$ for every truth assignment $v$ such that $v(A)=T$. But that's a theorem, not something that has to be true for any proof system. – David C. Ullrich Sep 03 '18 at 17:35
  • I guess I may need to look into exactly how an operator differs from a relation. I think I understand better now, but then what is $\models$ used for? I'm not sure I understand the $v(B) = T$ example, is $T$ here supposed to mean "true"? – user525966 Sep 03 '18 at 18:05
  • Probably I should just say yes, $T$ means "true". If we're being very careful $T$ and $F$ are just two arbitrary symbols that we manipulate according to certain rules. – David C. Ullrich Sep 03 '18 at 18:17
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    How an operator differs from a relation: They're not the same thing at all. For example, if I say "2+3" I haven't asserted anything; + is an operator, not a relation, so when I say "2+3" I've just pronounced a name for the number 5. On the other hand, $<$ is a relation: If I say "$2<3$" I have asserted something. – David C. Ullrich Sep 03 '18 at 18:20
  • Would it make sense to think of an operator like a function? Takes some inputs, returns an output? (although I don't know if this is a circular definition since functions rely on set theory and then set theory relies on logic, etc) – user525966 Sep 03 '18 at 18:35
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    Yes, an operator is exactly a function! Except of course for the notation; since $+$ is a binary "infix" operator we write $2+3$ instead of $+(2,3)$. (Yes, it seems there's always some circularity. The formal definitions in logic talk about sets of this and sets of that - then formal set theory is defined using formal logic...) – David C. Ullrich Sep 03 '18 at 18:48
  • Sort of like the sets referenced here? https://en.wikipedia.org/wiki/Propositional_calculus#Generic_description_of_a_propositional_calculus – user525966 Sep 03 '18 at 19:02
  • Sometime A → B may seems to be an assertion, when it appears in something like Γ ⊢ A → B, which may explain the confusion. – Hibou57 Sep 18 '20 at 05:48
  • @user525966, x - y > 0 is a formula which reduces to the boolean true or false, it may reduce to true as much as it may reduce to false, the same with A → B. The key word here “reduce to”, you may substitute what the expression reduces to for the expression (ex. if x - y > 0 is false, you may replace x - y > 0 by false). While A ⊢ B is not intended to be reduced, if you would replace A ⊢ B by ⊤, it would just be useless. – Hibou57 Sep 18 '20 at 06:22
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We have in place a very sharp distinction between the object language connective $\to$ and the metalinguistic sign $\vdash$ for the derivability relation (and the metalinguistic sign $\vDash$ for the logical consequence (or entailment) relation).

$(P \land Q) \vdash Q$ expresses the existence in the propositional claculus of an argument.

The metalinguistic formula asserts that we have a derivation of $Q$ from hypothesis $P \land Q$.

A derivation in the calculus is the formal counterpart of the concept of inference: every step in the derivation is the application of a rule of inference (like e.g. modus ponens) and a rule of inference is the formalization of an "elementary step" in the inferential process.

The formula $(P \land Q) \to Q$ is a single formula in the language of propositional calculus.

If we assert it, we are assering that "either $(P \land Q)$ is false or $Q$ is true".

In ordwer to appreciate the difference, we have to consider that we can formalize the propositional calculus with only teo conenctives :

$\land$ and $\lnot$ (or $\lor$ and $\lnot$)

but the derivability relation does not change its definition.

Of course, there is a link between the two notions, and the link is formalized by the meta-logic property of the calculus expressed by the Deduction Theorem stating that :

if a formula $B$ is derivable from a set of assumptions $\Delta \cup \{A\}$, then the formula $A \to B$ is derivable from $\Delta$.

The deduction theorem is a formalization of the common proof technique in which an implication $A \to B$ is proved by assuming $A$ and then deriving $B$ from this assumption conjoined with known results.

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Mauro ALLEGRANZA
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  • But say we only formalize the calculus with $\lnot, \land$. How do we then define and show how some new symbol like $\to$ works? – user525966 Sep 03 '18 at 14:55
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    @user525966 - we introduce $p \to q$ as an abbreviation of the formula $\lnot p \lor q$ (in classical logic) or as an abbreviation for $\lnot (p \land \lnot q)$. – Mauro ALLEGRANZA Sep 03 '18 at 15:04
  • But even after we introduce that symbol with that definition, how do we then know it is the appropriate connective to describe movement in things like mathematical proofs? How do we define how $\land$ and $\lnot$ work without truth tables? Do we have to define true and false? – user525966 Sep 03 '18 at 15:39
  • @user525966 - two steps. Syntactically, connectives are "defined" in thecalculus through the axioms and rules. Then, semantically, we have the truth tables for the connectives. We want that the two approaches match: this is the role of soundness and completeness properties: we want that the axioms and theorems of the calculus are sound (i.e. TRUE according to the semantical interpretation) and complete (i.e. that all TRUE formuklas will be derivable). – Mauro ALLEGRANZA Sep 04 '18 at 06:19
  • Are truth tables the only way to give semantic quality? I thought they were more of a convenience rather than a necessity. The semantics can't be extracted from the way the axioms/inference rules are set up? Are truth tables metalogical? – user525966 Sep 04 '18 at 12:07
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In our world, it is true that if it rains, things will get wet. We can express this truth as:

$R \rightarrow W$

But it is certainly not true that:

$R \vdash W$

That is, given $R$, we can't logically infer $W$. Why? Because we can imagine worlds where nothing gets wet, even if it rains.

Bram28
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  • So $A \vdash B$ means $B$ is provable from $A$ regardless of the system we use, regardless of the "world" we're in? – user525966 Sep 03 '18 at 17:26
  • Saying we can express this truth as $A\to B$ is really totally missing the point. – David C. Ullrich Sep 03 '18 at 17:40
  • @user525966 What $A\vdash B$ means depends on what proof system we're talking about. A proof system has axioms and inference rules - given a proof system, saying $A\vdash B$ means there is a finite sequence of wffs $A_1,\dots, A_n$ such that $A_n=B$ and each $A_j$ is either $A$, an axiom, or a consequence of previous $A_i$ by one of the rules. That's all it means. – David C. Ullrich Sep 03 '18 at 17:46
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    Why I say it's missing the point: What you're actually explaining here is more the difference between $A\to B$ and $\models(A\to B)$. Of course it's a theorem that (for standard proof systems) $\models(A\to B)$ if and only if $A\vdash B$, but the two don't mean the same thing! If they meant the same thing the Soundness and Completeness theorems would have no content. – David C. Ullrich Sep 03 '18 at 17:59
  • @DavidC.Ullrich The $\vdash$ symbol then I imagine is mostly used in those inference rules in set $Z$ (I'm reading https://en.wikipedia.org/wiki/Propositional_calculus#Generic_description_of_a_propositional_calculus when I reference this), giving you the rules that dictate how to jump from one wff (or collection thereof) to another wff. And then axioms are basically inference rules that don't have any premises, i.e. the conclusions come out of nothing by default? – user525966 Sep 03 '18 at 18:08
  • Yes, an axiom is (equivalent to) an inference rule with no premises. At that link $Z$ is a set of inference rules, so I'm not sure exactly what "$\vdash$ is used in those inferene rules in set $Z$". A true fact, possibly what you mean, is that what $\vdash$ means depends on what set $Z$ we're considering. One might write $A\vdash_Z B$ to mean there is a proof of $B$ from $A$, using the rules in $Z$. People write just $A\vdash B$ instead because $Z$ is understood. – David C. Ullrich Sep 03 '18 at 18:27
  • @DavidC.Ullrich I refer to how examples 1 and 2 below (on that link) they use has modus ponens as the inference rule in $Z$ and modus ponens uses the $\vdash$ symbol, etc – user525966 Sep 03 '18 at 19:27
  • @user525966 Oh, I see what you mean. Alas this is another closely related use of the symbol $\vdash$, not quite the same as the sense we've been talking about. Instead of using the notation $A,A\to B\vdash B$ to define modus ponens, some people use the perhaps preferable notation $$\begin{aligned}&A\&A\to B\&_______\&B\end{aligned}$$. – David C. Ullrich Sep 03 '18 at 19:32
  • So that symbol is used in different ways? What's the difference between that and this big line notation, if it's not technically accomplishing the same thing? Yowza this logic stuff is super confusing. – user525966 Sep 03 '18 at 19:37
  • When $\vdash$ is used in defining an inference rule what it means iis the same as that horizontal line. Seems to me that using $\vdash$ in defining inference rules is a bad idea, since it's a slightly different meaning of $\vdash$ - we use $\vdash$ in defining $Z$ even though we need $Z$ to define $\vdash$ (in the sense we've been talking about here). – David C. Ullrich Sep 03 '18 at 19:46