112

Consider A $\Rightarrow$ B, A $\models$ B, and A $\vdash$ B.

What are some examples contrasting their proper use? For example, give A and B such that A $\models$ B is true but A $\Rightarrow$ B is false. I'd appreciate pointers to any tutorial-level discussion that contrasts these operators.

Edit: What I took away from this discussion and the others linked is that logicians make a distinction between $\vdash$ and $\models$, but non-logicians tend to use $\Rightarrow$ for both relations plus a few others. Points go to Trevor for being the first to explain the relevance of completeness and soundness.

YuiTo Cheng
  • 4,705
user287424
  • 1,239
  • There are by now many similar q's on math.se: http://math.stackexchange.com/questions/90787/whats-the-difference-between-to-implication-and-vdash-therefore/90801#90801, http://math.stackexchange.com/questions/68932/whats-the-difference-between-material-implication-and-logical-implication?rq=1 and many more – alancalvitti Jan 24 '13 at 21:00
  • 1
    I'd like to highlight a problem when you misuse them you get problematic results, or at least I did: https://math.stackexchange.com/questions/854752/can-a-statement-in-fol-be-equivalent-to-two-separate-independent-statements – dezakin Jul 04 '14 at 17:57

7 Answers7

86

First let's compare $A \implies B$ with $A \vdash B$. The former is a statement in the object language and the latter is a statement in the meta-language, so it would make more sense to compare $\vdash A \implies B$ with $A \vdash B$. The rule of modus ponens allows us to conclude $A \vdash B$ from $\vdash A \implies B$, and the deduction theorem allows to conclude $\vdash A \implies B$ from $A \vdash B$. Probably there are exotic logics where modus ponens fails or the deduction theorem fails, but I'm not sure that's what you're looking for. I think the short answer is that if you put a turnstile ($\vdash$) in front of $A \implies B$ to make it a statement in the meta-language (asserting that the implication is provable) then you get something equivalent to $A \vdash B$.

Next let's compare $A \vdash B$ with $A \models B$. These are both statements in the meta-language. The former asserts the existence of a proof of $B$ from $A$ (syntactic consequence) whereas the latter asserts that every $B$ holds in every model of $A$ (semantic consequence). Whether these are equivalent depends on what class of models we allow in our logical system, and what deduction rules we allow. If the logical system is sound then we can conclude $A \models B$ from $A \vdash B$, and if the logical system is complete then we can conclude $A \vdash B$ from $A \models B$. These are desirable properties for logical systems to have, but there are logical systems that are not sound or complete. For example, if you remove some essential rule of inference from a system it will cease to be complete, and if you add some invalid rule of inference to the system it will cease to be sound.

Trevor Wilson
  • 16,989
  • 15
    It would be wise to discern between $\Rightarrow$ (logical implication) and $\rightarrow$ (logical connective). – Asaf Karagila Jan 24 '13 at 19:44
  • @Asaf I'm not sure I know the difference... – Trevor Wilson Jan 24 '13 at 19:45
  • From the textbook presentation, I feel somewhat clear about where each operator is conventionally used. However, I feel that I must be missing a lot, because I can't think of examples where substituting one of these for another would actually make a true statement false. – user287424 Jan 24 '13 at 19:47
  • 9
    The former is a statement about sentences; the latter is a part of a sentence. Logical implication vs. material implication. – Asaf Karagila Jan 24 '13 at 19:48
  • well, substituting $A \implies B$ for $A \vdash B$ or vice versa would generally make a true statement ungrammatical. But to have one of the things you mention be true while the other is false would require considering some unusual logical systems. In the second paragraph I mention one way to get them. – Trevor Wilson Jan 24 '13 at 19:50
  • @Asaf I see. I was reading the $\implies$ in the question as the logical connective of material implication. Isn't logical implication what is expressed by the turnstile? – Trevor Wilson Jan 24 '13 at 19:52
  • Trevor, I gathered that much. But it is common to use $\implies$ for logical, rather than material. The turnstile is syntactical and $\implies$ is semantic; but so is $\implies$. – Asaf Karagila Jan 24 '13 at 20:00
  • @AsafKaragila, Wikipedia and others sometimes write statements in a column, followed by a horizontal line and then a conclusion, eg http://en.wikipedia.org/wiki/Modus_ponens. What is the horizontal line, implication, connective, or something else? – alancalvitti Jan 24 '13 at 20:09
  • @TrevorWilson, but isn't also $A \to B$ an assertion that it is provable? – alancalvitti Jan 24 '13 at 20:25
  • @alancalvitti I think the horizonal line is the same as the turnstile in that context – Trevor Wilson Jan 24 '13 at 20:32
  • @alancalvitti Not all true statements are provable (but one doesn't need Goedel's incompleteness theorem to see that a statement is not exactly the same thing as the assertion of the provability of that statement.) – Trevor Wilson Jan 24 '13 at 20:34
  • @TrevorWilson, agreed, even the Greeks distinguished theorem and proof. But for example, is Modus Ponens provable? – alancalvitti Jan 24 '13 at 20:48
  • @TrevorWilson, how would you prove "B holds in every model of A" if the list of models is open-ended or potentially unbounded? – alancalvitti Jan 24 '13 at 20:51
  • 1
    @alancalvitti Modus Ponens is a valid rule of inference in many logics. I think we're getting off-topic here, but see https://en.wikipedia.org/wiki/Modus_ponens – Trevor Wilson Jan 24 '13 at 20:59
  • 1
    @alancalvitti By saying "Let $M$ be a model of $A$.... Therefore $B$ holds in $M$" where the stuff in between depends on the situation. – Trevor Wilson Jan 24 '13 at 21:01
  • @TrevorWilson, yes I already linked to that in a comment above. What do you mean by "valid"? You mean Provable? - It's on-topic: look at the title of the Q. – alancalvitti Jan 24 '13 at 21:01
  • @TrevorWilson "where the stuff in between depends on the situation" I hope you're joking. This is closer to writing fiction than math. – alancalvitti Jan 24 '13 at 21:02
  • @AsafKaragila, can you please help? From this and similar Q's posted on math.se there is little consistency in how people interpret these concepts. – alancalvitti Jan 24 '13 at 21:02
  • @alancalvitti Ah, I even read the page and responded to the comment. But I forgot that you had linked to it, sorry. What it means for a rule of inference to be valid is relevant to the question, I think. – Trevor Wilson Jan 24 '13 at 21:03
  • @alancalvitti I'm not joking. Different statements have different proofs. I can't prove $A \models B$ without knowing what $A$ and $B$ are. – Trevor Wilson Jan 24 '13 at 21:04
  • @TrevorWilson, can you prove $A \to B$, $A \implies B$ or $A \vdash B$ without knowing what $A$ and $B$ are? Or maybe you're saying some of these are not accompanied by proofs? – alancalvitti Jan 24 '13 at 21:09
  • 2
    @alan: I am writing from my iPhone, I am currently busy with marinating my brain in alcohol in the local watering hole. I will help you (if I can) later. – Asaf Karagila Jan 24 '13 at 21:24
  • 1
    @Asaf, I always pictured you as some crusty, aged Hasidic type but I see on your web site you're a PhD candidate - ok drink to many more decades of math.. Ps, since you're interested in sets w/o Choice, funny comment from George Bergman: he told me MacLane told him that Berkeley mathematicians "worship at the temple of ZFC" and he took offense to that. But why? – alancalvitti Jan 25 '13 at 02:28
  • 1
    @alan: I am as far from a Hassidic Jew as possible. As for why he took offense? I don't know. Ask him yourself. – Asaf Karagila Jan 25 '13 at 09:59
  • Is the horizontal line (see http://en.wikipedia.org/wiki/Simply_typed_lambda_calculus#Typing_rules) always a turnstile? Why do they use the horizontal line? Is it just for formatting? If the horizontal line is a turnstile (|-) why do they have more turnstiles on the top and bottom of the horizontal line? – CMCDragonkai Dec 29 '14 at 10:26
  • @CMCDragonkai I don't know much about the lambda calculus, but I think the horizontal line and turnstile are used in it to mean different things. (Also, the turnstile might mean a different thing in lambda calculus than it does in predicate logic, as far as I know.) – Trevor Wilson Dec 29 '14 at 22:15
  • In other places I've read, they denote the top of the horizontal line as the condition, and the bottom as the result. It sounds like a meta statement of the language. Saying, if this can be proved, then the bottom can be proved? That sounds like a turnstile to me... – CMCDragonkai Dec 30 '14 at 04:31
  • @CMCDragonkai I wasn't saying that the horizontal line in lambda calculus means something different than the turnstile in predicate logic. Indeed, they seem to mean something similar. But I couldn't say for sure. – Trevor Wilson Dec 31 '14 at 18:55
  • Maybe somebody want to expand https://en.wiktionary.org/wiki/%E2%8A%A8 – Vladimir Reshetnikov Feb 23 '18 at 18:22
  • I have a doubt about the relation between ⊢ and ⇒, I’m unsure ⇒ can just be a syntactic consequence, rather a semantic consequence, since P ⇒ Q means P → Q is unconditionally True. Unless I am missing something … – Hibou57 May 03 '20 at 17:00
  • 2
    @Hibou57 I have seen the symbol $\implies$ used to mean different things. I was taking it to be the logical connective of material implication, which some people instead call $\to$, because that was how I interpreted its use in the OP's question. If I hadn't been trying to match OP's terminology, all instances of $\implies$ in my answer would be $\to$ instead. – Trevor Wilson May 03 '20 at 20:13
51

@Trevor's answer makes the crucial distinctions which need to be made: there's no disagreement at all about that. Symbolically, I'd put things just a bit differently. Consider first these three:

$$\to,\quad \vdash,\quad \vDash$$

  1. '$\to$' (or '$\supset$') is a symbol belonging to various formal languages (e.g. the language of propositional logic or the language of the first-order predicate calculus) to express [usually, but not always] the truth-functional conditional. $A \to B$ is a single conditional proposition in the object language under consideration.
  2. '$\vdash$' is an expression added as useful shorthand to logician's English (or Spanish or whatever) -- it belongs to the metalanguage in which we talk about consequence relations between formal sentences. Unpacked, $A, A \to B \vdash B$ says in augmented English that in some relevant deductive system, there is a proof from the premisses $A$ and $A \to B$ to the conclusion $B$. (If we are being really pernickety we would write '$A$', '$A \to B$' $\vdash$ '$B$' but it is always understood that $\vdash$ comes with invisible quotes.)
  3. '$\vDash$' is another expression added to logician's English (or Spanish or whatever) -- it again belongs to the metalanguage in which we talk about consequence relations between formal sentences. And e.g. $A, A \to B \vDash B$ says that in the relevant semantics, there is no valuation which makes the premisses $A$ and $A \to B$ true and the conclusion $B$ false.

As for '$\Rightarrow$', this -- like the informal use of 'implies' -- seems to be used (especially by non-logicians), in different contexts for any of these three. It is also used, differently again, for the relation of so-called strict implication, or as punctuation in a sequent. So I'm afraid you do just have to be careful to let context disambiguate. The use of the two kinds of turnstile is absolutely standardised. The use of the double arrow isn't.

Peter Smith
  • 54,743
  • 3
    I should point out that in Israel you see $\implies$ used for $\models$ all the time, even in basic logic courses. The $\models$ is a relation between a structure and a sentence, rather than two formulas. – Asaf Karagila Jan 24 '13 at 20:08
  • 1
    (1) Doesn't $A \to B$ also mean that there is no valuation which makes the premise $A$ true and the conclusion $B$ false? (2) Why isn't Modus Ponens also a sentence rather than a relation, namely: "If $A$ and $A \to B$ is true, then $B$ is true." – alancalvitti Jan 24 '13 at 20:23
  • 4
    @AsafKaragila Interesting. In my neck of the words, $\vDash$ gets a double use, for the relation of semantic entailment between wffs (or, officially, a set of wffs) and a wff, as in $\Gamma\vDash\varphi$, and [not unconnectedly of course] also for the modelling relation $\mathfrak{A}\vDash\varphi$. And mathmos often use $\Rightarrow$ for the conditional! – Peter Smith Jan 24 '13 at 20:27
  • 1
    @alancalvitti (1) A isn't a premiss in $A \to B$. Premisses of arguments are typically asserted; $A \to B$ never asserts the antecedent $A$. (2) Modus ponens is a rule, not a proposition (as Lewis Carroll long ago reminded us!). – Peter Smith Jan 24 '13 at 20:30
  • @PeterSmith, do the following terms (which all appear on this page) have mathematical definitions? "Premiss" "Rule" "Sentence" "Formula" "Structure" "Provable" – alancalvitti Jan 24 '13 at 20:45
  • @alancalvitti Standard logical textbooks give definitions. – Peter Smith Jan 24 '13 at 20:55
  • 1
    @PeterSmith, but they're typically defined in terms of similar terminology rather than rigorously. So it is difficult to determine their mutual consistency. For example, what does "Sentence" mean? – alancalvitti Jan 24 '13 at 20:57
  • 1
    A sentence is a formula with no free variables. – Benedict Eastaugh Feb 16 '13 at 11:29
  • $\vdash$ can also be used to denote sequents (and confusingly $\rightarrow$ and $\Rightarrow$ is also used) and typing judgements. The notation truly is a mess. – Poscat Mar 21 '24 at 12:50
17

$A\models B$ means $B$ is true in every structure in which $A$ is true.

$A\vdash B$ means $B$ can be proved if $A$ is assumed. But what is a proof? Usually one wants to define "proof" in such a way that (1) there's an algorithm for deciding which putative proofs are really proofs; and (2) if $A\vdash B$ then $A\models B$, i.e. only those things are provable that ought to be. In many reasonable circumstances, one also has: if $A\models B$ then $A\vdash B$, i.e. everything that ought to be provable is provable.

10

An additional comment, motivated by Peter Smith's explanation :

'$\to$' (or '$\supset$') is a symbol belonging to various formal languages (e.g. the language of propositional logic or the language of the first-order predicate calculus) to express [usually, but not always] the truth-functional conditional. $A \to B$ is a formula of object language under consideration with the conditional connective as main operator.

"Implies" symbols (both the syntactical one : '$\vdash$' and the semantical one: '$\vDash$' ) are used in the meta-language to express the consequence relation.

The connection between them is established by the rule of Modus Ponens that allows us to conclude $A \vdash B$ from $\vdash A \supset B$ , and by the Deduction Theorem that allows us to conclude $\vdash A \supset B$ from $A \vdash B$.

Finally, the Completeness Theorem establish the connection betwenn '$\vdash$' and '$\vDash$'.

So there is a very tight link between the three symbols, but they must be treated as distinct.

It is useful to think that it is possible to avoid '$\supset$' as connective (see Whitehead & Russell's Principia Mathematica (1910)) for example using only negation ($\lnot$) and disjunction ($\lor$).

7

As @Trevor Wilson said, $\vdash$ which is named “turnstile” or “right tack”, belongs to the meta‑language, however, it's not always a syntactic consequence operator, it also is a semantic consequence (what @Trevor Wilson said is $\models$), at least in type theory. The model is not only ⊨ , U+22A8, named TRUE in the Unicode database, it's also ⊧ , U+22A7, precisely named MODELS, in the Unicode database. TeX and Unicode disagree here (about ⊨ and ⊧).

As @Peter Smith said, $\rightarrow$ is a functional implication. I'm adding it's related to $\supset$ in the way the former is an interpretation of the latter (one domain to another domain).

Examples

$$\frac{\Gamma (x) = \tau}{\Gamma \vdash x : \tau}$$

The above is from type theory, $\vdash$ stands for a semantic consequence, something which can't be inferred from the syntax. In this example, $\Gamma$ stands for a typing context, which is a semantic context, not a syntactic context; there, $\Gamma$ is like a function, something computed at the meta‑level.

$$A \supset B = A \rightarrow B$$

The above is from Curry‑Howard correspondence. To make the above clearer, it appears in a context which also asserts this:

$$A \land B = A \times B$$

… where $A \times B$ is a term of functional language and lambda‑calculus, so is $A \rightarrow B$. $A \rightarrow B$, belongs to the domain of $A \times B$; and $A \Rightarrow B$ or $A \supset B$, belongs to the domain of $A \land B$.

This is the same as the previous above (same interpretation holds):

$$A \Rightarrow B = A \rightarrow B$$

The distinction is less strong than object‑level / meta‑level distinction, this is a domain distinction. Comparing “$\vdash$ vs $\Rightarrow$” to “$\Rightarrow$ vs $\rightarrow$” is a bit like comparing explicit type conversion to implicit type conversion… a comparison to be used with a lot of care, that's just to give a picture.

Counter examples (as the OP wish)

$$A \rightarrow B $$ Where a context is $\Gamma$, the above may make sense.

$$A \vdash B $$ Where a context is $\Gamma$, the above makes no‑sense at all (means nothing).

$$A \Rightarrow B$$ $$A \rightarrow B$$ In typed lambda‑calculus (simply typed or more expressive), both of the above may be seen as the same (the latter is originally an interpretation of the former, but since, in a valid type theory the reverse interpretation is also valid). In untyped lambda‑calculus, the former do not exists, and the latter may make sense, as much as it may makes no sense.

$$\vdash A $$ You may see the above, this makes sense (to be read as “derivable from the empty context)”.

$$\rightarrow A $$ $$\Rightarrow A $$ $$\subset A $$ You will never see the above, this makes no sense.

May be completed with other examples and counter‑examples, a future day.

Hibou57
  • 377
  • 4
    I learned quite a bit from this even though I didn't ask the question. Although I will point out that statements like A⇒B=A→B should have brackets because = takes the same precedence as a lot of other logical operators. There is no order of operations for logic that is agreed upon. Although I am nick-picking a bit because everyone can understand what you wrote anyway. – user400188 Jan 22 '17 at 06:41
5

A $\vdash_{i.e.}$ B : (read B can be proven by i.e. using A) means algorithm "ie", an inference engine, can obtain B from A

A $\models$ B : (read A entails B) means statement B is true in all states of the world, given A is true.

Specifically, if we let A be the axioms of some system. Then:

  1. The system is complete if some possible algorithm can prove all the true statements in the system. In other words, If B is true given A (axioms) means there exists some algorithm "i.e" that can prove B using A
  2. The system is sound if no statement that is false in any state of the world can be proven by any inference engine given the axioms of the system. In other words, if some algorithm "i.e." proves B using A (axioms), then B is true, given A.
Rahul Madhavan
  • 2,789
  • 1
  • 11
  • 14
4
  • material conditional $\left(\to\right)$
  • implication$\left(\Rightarrow\right):$
    $\quad$ is true in the given interpretation and axiom system and context
  • logical implication / (semantic) logical entailment $\left(\models\right):$
    $\quad$ is true regardless of interpretation
  • derivability / syntactic logical entailment $\left(\vdash\right):$
    $\quad$ can be proven true regardless of interpretation

(, , are metalanguage symbols, while is in the object language.)

For example, these two claims are simultaneously plausible: \begin{align}&\forall x\;\; x=x &\Rightarrow &&\forall x\,\forall y\;\;\; x^2 -y^2 = (x+y)(x-y),\\&\forall x\;\; x=x &\not\models &&\forall x\,\forall y\;\;\; x^2 -y^2 = (x+y)(x-y) .\end{align}

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

P.P.S. To be clear: although I distinguish analytical and synthetic implication from logical entailment , in practice I do frequently use (which is better recognised) even when I specifically mean the latter.

ryang
  • 38,879
  • 14
  • 81
  • 179
  • What is meant by "interpretation" here? I understand that ⇒ is a meta-proposition about a "→"-statement, basically saying "$( \forall x, ( p(x) \rightarrow q(x) ) ) = True$. So, is ⊨ even more meta? What is the context (by Hibou57) the "in all states of the world" (by Rahul Madhavan) or "in every structure" (by Michael Hardy)? I probably misunderstand if I think this means "no matter the axioms", because they are an integral part of the language I am using in the first place? Also, specifically: Why is $∀x(x=x)⊭∀x∀y(x2−y2=(x+y)(x−y))$? – Make42 May 19 '22 at 13:07
  • @Make42 $;{^2},,+,,-,,()(),$ are not logical operators; if I define $x^2:=x$ and $x+y:=x$ and $(x)(y):=x$ and $x-y:=y,$ then that statement has a true antecedent and false consequent. On the other hand, that statement is mathematically true. – ryang May 19 '22 at 15:18
  • So, does this mean that 1) if a claim which uses ⊨ is true then the same claim using ⇒ instead is also true? 2) once I use symbols that are not logical symbols, then I cannot use ⊨ anymore, but - given the interpretation, I might still be able to use ⇒? – Make42 May 19 '22 at 19:24
  • @Make42 1. Yes. 2. No: 1=1 ⊨ 0+0=0+0 is correct. – ryang May 19 '22 at 19:31
  • I did as you asked and wrote my own question, even though it seems to be a mess. – Make42 May 20 '22 at 20:04