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Why do abstract algebra texts generally define a group something like more-or-less this...

Let * denote a binary operation on a set G.

For all x, y, z in G x*(y*z)=(x*y)*z

There exists an element 1 in G, such that for all x in G, x*1=x

For all x in G, there exists an x' in G, such that x*x'=1.

Instead of say using a definition like this:

Let * denote a binary operation, ' a unary operation, 1 a nullary operation on G (as best as I understand it, a nullary operation can get thought of a mapping from the empty set to G here, since it does hold that "if for all x belonging to the empty set, then there exists exactly one output in G", since the hypothesis in quotes comes as false, this statement always obtains.)

For all x, y , z in G x*(y*z)=(x*y)*z

For all x in G, x*1=x

For all x in G, x*x'=1

I know that S. C. Kleene's Mathematical Logic has a short section where a small few theorems of group theory get proved using something like the second definition of a group, but only a handful, and little else of group theory or any other algebra gets developed. Does anyone know of texts that study some algebraic systems like groups, rings, monoids etc. using definitions more like the second than the first? That is, study algebras not like how universal algebra texts just mention (maybe I haven't read this thoroughly enough though) several algebras and move on, but rather actually study particular algebras using definitions like the second? Additionally, I see that the Schaum's Outline of Group Theory happily uses a fair amount of reverse Polish notation. Does anyone know of any algebra texts that use either a fair amount of Polish notation, or a fair amount of reverse Polish notation?

Thanks for any help here.

Doug Spoonwood
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    A nullary operation on $G$ is, by definition, a map from $G^{\emptyset}$ to $G$. By definition, $G^{\emptyset}$ is the set of all functions from the empty set to $G$; the only function from the empty set to $G$ is $\emptyset$, so a nullary operation on $G$ is a function with domain ${\emptyset}$ and codomain $G$. The final effect is that a nullary operation is "equivalent" to specifying a distinguished element of $G$ (the image of the unique element of $G^{\emptyset}$). \Tour description is missing a 'mapping'; should be a mapping from the set of mappings from the empty set to G, to G – Arturo Magidin May 28 '11 at 03:01
  • What's your definition of a function? I define a function simply as a mapping from all members of the source set to exactly one member of the target set. In other words, if s belongs to the source set S, there exists a unique element t belonging to the target set T such that the function F takes s from S to some unique t belonging to T. Or equivalently, if s belongs to S, then F takes s to at least one t in T, and if F takes s to t' in T also, then t=t'. Under this definition, every conceivable way of mapping from the empty set comes as permissible, why do you introduce a restriction? – Doug Spoonwood May 28 '11 at 04:10
  • I'll add that nullary operations seem to get thought of as constants also. A set of just numbers (reals, rationals, integers, naturals, complex numbers) consists of a set of nothing but constants. So, if you think of a nullary operation as a constant, and a set consists of a collection of constants (as they at least proto-typically do), then every element of the target set of a function should qualify as a nullary operation. The definition I've given of function allows exactly for this, since the hypothesis comes as false, you can infer whatever you please for nullary operations. – Doug Spoonwood May 28 '11 at 04:25
  • Also, the definition of a function comes as a map from the Cartesian product space SxSx...xS to some set T, then if sxsx...xs belongs to SxSx...xS, then there exists a unique member of T mapped from SxSx...xS to T. So, if we have no sets S, then the hypothesis of sxsx...s belonging to SxSx...xS comes as false. So, a nullary operation thus allows for every element of a set to qualify as a nullary operation, as well as a single unique element of nullary operation. We should get exactly such behavior if the empty set loosely corresponds to the "false" of classical logic. – Doug Spoonwood May 28 '11 at 05:00
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    @Doug: See here - for any set $A$, there is only one function from the empty set to $A$. As Arturo correctly explains, a nullary operation can be thought of as a function from "$G^0$" to $G$, where "$G^0$" denotes "$G$ cross itself 0 times", a concept which is not exactly well-defined, but which is made rigorous by the identification of $G^n$ with the collection of functions from an $n$-element set $X$ to $G$. – Zev Chonoles May 28 '11 at 05:32
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    In case you are not familiar with it, the notation $Y^X$ refers to the set of functions from $X$ to $Y$. Since $G^0=G^{\varnothing}$ is a singleton (consisting of the unique function from $\varnothing$ to $G$), the functions from $G^0$ to $G$ are in bijection with the elements of $G$, by sending a function $f:G^0\rightarrow G$ to its image $f(G^0)$, and conversely, sending an element $g$ of $G$ to the unique function $f$ sending the sole element of $G^0$ to $g$. – Zev Chonoles May 28 '11 at 05:36
  • The answer to questions of this form is almost always "historical inertia." – Qiaochu Yuan May 28 '11 at 09:36
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    @Doug: I introduce no restrictions, but you are getting yourself all confused. Two functions are equal if and only if they have the same domain, the same codomain, and the same value at every point of the domain. As such, there is one and only one function from the empty set to $X$, for each set $X$. "Every conceivable way of mapping" comes out to just one function. For cartesian products, the cartesian product $\pi_{i\in I}X_i$ is defined to be the set of all functions from the index set to the union of the $X_i$, with the value of $i$ in $X_i$. (cont) – Arturo Magidin May 28 '11 at 11:33
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    @Doug: (cont) So the set $G^0$ is, by definition the set of all functions from $0=\emptyset$ to $G$. As just explained, there is one and only one functions. For an ordinal $\alpha$, an $\alpha$-ary operation on $X$ is a function $X^{\alpha}\to X$, so a nullary operation is a function $X^0\to X$. Since $X^0={\emptyset}$, a nullary operation on $X$ is a function ${\emptyset}\to X$; equivalently, a single element of $X$. – Arturo Magidin May 28 '11 at 11:35

5 Answers5

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In fact, if you want to define groups as a variety of $\Omega$-algebras, one does in fact define a group this way: as an algebra with signature $(2,1,0)$ and so on.

This lets you fit group theory (and later, ring theory) into the wider tapestry of Universal (or General) Algebra; see for example George Bergman's An Invititation to General Algebra and Universal Constructions.

However, as it happens, there are a lot of things which are true for groups that are not generally true for universal algebras. To consider one example, any semigroup homomorphism between two groups must in fact be a group homomorphism. This is not the case even for monoids (you can have a semigroup homomorphism between two monoids that is not a monoid homomorphism, because it does not map the identity to the identity). If you don't know groups "well enough" (and view them merely as universal algebras with signature $(2,1,0)$ and satisfying appropriate identities), then in order to check that a map between groups is a homomorphism you would need to check that it maps products to products, inverses to inverses, and the identity to the identity. In order to "get away" with just checking that it is a semigroup homomorphism, you would need to prove that this is the case... and this turns out to be essentially equivalent to performing the verification that the definition of group you quote first uniquely determines the identity and the inverses.

So, if they are equivalent, why use one and not the other? A couple of reasons: one is certainly historical inertia. Groups were originally defined only in terms of their binary operation. The second is parsimony: when you have a concept, and it's ubiquituous (as the group concept is) you want your definition to be as parsimonious as possible, because you want to make it easy to verify that a given instance is in fact a group. We could add all sorts of clauses to the definition of groups (clauses which are theorems of the standard definition) that would make further theorems much easier to prove; but that would mean that if you find a set lying on the street and you want to check if it is a group, you would need to check all the extra clauses.

Using the "usual" definition, you only need one operation and three properties. Using the universal algebra definition, you would need to check three operations, and three properties. So you end up having to do more checking to see if what you have before you is indeed a group or not. Generally, it's better to need to do less checking rather than more checking to see if the theory applies.

I don't really know any text that use prefix or suffix notation for the binary operation of a semigroup, so I can't help with the final question.

Since this came up in the comments, let's clarify a few things about "nullary operation."

First, whether you define a function as a set of ordered pairs, or you define it in terms of a domain, a codomain (and regardless of how you define codomain), and a rule that associates to every element of the domain a single element of the codomain, I think we can agree that if $f$ and $g$ are functions with the same domain and same codomain, then $f=g$ if and only if for every element $x$ in the domain, $f(x)=g(x)$.

So, if $A$ is a set, how many distinct functions are there from $\emptyset$ (the emptyset) to $A$? As you note, the condition of being a function (for every element of the domain there is one and only one element of $A$) is satisfied vacuously, so it seems like you can just take your function to be "whatever", and it will satisfy the definition.

But the point I raised in comments was that in point of fact, there is only one function because of what equality of functions means. If $f$ and $g$ are two functions with domain $\emptyset$ and codomain $A$, then $f=g$ by vacuity: for every $x\in\emptyset$, we have $f(x)=g(x)$. (Put another way, for $f$ and $g$ to be different, there would have to be an element in the empty set where $f$ and $g$ disagree; no such element, so they aren't different). That means that any two functions with domain $\emptyset$ and codomain $A$ are necessarily equal as functions. So there is one and only one function with domain $\emptyset$ and codomain $A$.

If you define a function from $X$ to $Y$ as a subset of $X\times Y$ (ordered pairs) that satisfies certain properties, then it turns out that $\emptyset$, as a function from $\emptyset$ to $A$ (as a subset of $\emptyset\times A = \emptyset$) satisfies this definition, so the function from $\emptyset$ to $A$ is $\emptyset$, the "empty function." That's why I said that there is one and only one function from the empty set to $A$.

Now, about $n$-ary (and $0$-ary or nullary) operations: generally, an $n$-ary operation on $A$ is defined to be a function from $A^n$ to $A$. So a nullary operation is a function with domain $A^0$ and codomain $A$.

What is $A^0$? Well, here it is useful to define arbitrary cartesian products: if $\{A_i\}_{i\in I}$ is a family of sets, then $\mathop{\times}\limits_{i\in I}A_i$ is defined to be the set of all functions $f\colon I\to \cup A_i$ such that $f(i)\in A_i$. For $I=\{1,2,\ldots,n\}$, this can be naturally seen to be "the same" as the idea of $n$-tuples: the $n$-tuple $(a_1,\ldots,a_n)$ is identified with the function that maps $i$ to $a_i$; and a function $f\colon I\to \cup A_i$ with $f(i)\in A_i$ can be associated with the tuple $(a_1,\ldots,a_n)$.

So instead of defining $A^n$ as the set of $n$-tuples, it makes more sense to define it as the product $\mathop{\times}\limits_{i=1}^n A$; this allows for easy generalization to other sets: for any set $I$, you can define an $I$-tuple of elements of $A$ simply by $\mathop{\times}\limits_{i\in I} A$. This is $$\mathop{\times}_{i\in I} A = \bigl\{ f\colon I\to A\bigm| f\text{ is a function}\bigr\}$$ the set of all functions from $I$ to $A$. By analogy to $A^n$, we write this as $A^I$.

Using this notation, $A^0$ is the set of all functions from $0$ to $A$; under the usual definition of the natural numbers as sets, we have $0=\emptyset$, $1=\{0\}$, $2=\{0,1\}$, etc. So $A^0$ is $$A^0 = \{ f\colon 0 \to A\mid f\text{ is a function}\} = \{f\colon\emptyset\to A\mid f\text{ is a function}\}.$$ But we just talked about this. There is one and only one function with domain $\emptyset$ and codomain $A$; so $$A^0 = \{f\colon \emptyset\to a\mid f\text{ is a function}\} = \{\emptyset\}.$$

So, a nullary operation on $A$ is a function $A^0\to A$. But $A^0 = \{\emptyset\}$. So a nullary operation on $A$ is a function $\{\emptyset\}\to A$. There is a natural correspondence between functions from a singleton to $X$ and the elements of $X$: a function $f\colon\{a\}\to X$ corresponds to the element $f(a)$; and an element $b\in X$ corresponds to the function that sends $a$ to $b$. So the nullary operations on $A$ are in one-to-one correspondence with the elements of $A$, which is why nullary operations are sometimes called "distinguished elements of $A$": you can think of a nullary function as its unique value.

In any case, a nullary operation is not a mapping form the empty set to $G$, but rather it's a mapping from the set of all mappings from the empty set to $G$, to $G$: it's a function $G^{\emptyset}\to G$.

Arturo Magidin
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  • Thanks for the information and clarification! Sorry, I think I misread what you wrote above and thought you implied that there exists only one nullary operation in general. My mistake! Thanks also to Zev! – Doug Spoonwood May 29 '11 at 01:37
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    @Doug: I'm glad it was useful. – Arturo Magidin May 29 '11 at 01:42
  • But of course, since nullary operations come in one-to-one correspondence with the elements of A, there exist many possible in general. I think one might say that there exists only one nullary function in an "abstract" sense, similar to how there exists only one abstract trivial group, even though one can select any of a slew of elements for a concrete trivial group. A set with n elements has n nullary operations. – Doug Spoonwood May 29 '11 at 01:47
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    @Doug: No, again you are getting confused. A nullary operation is not a function with domain the emptyset, just like a binary operation is not a function with domain ${1,2}$. An $n$-ary operation has, as domain, a set of function, namely, the functions from $n$ to $A$. A nullary operation has as domain the set $A^0$, which is not empty, it's a set whose only element is the empty set (a bag that contains an empty bag is not itself empty). There is only one function from the empty set to $A$. But you cannot say that "there exists only one nullary function in the 'abstract sense'". – Arturo Magidin May 29 '11 at 01:49
  • Since the nullary operations come in one-to-one correspondence with the elements of A, since every operation qualifies as a function (but not conversely) also, there exist as many nullary functions as there exist elements of the set A. For example, we have three nullary functions for A={1, 2, 3}. So, either you have a contradiction here by claiming there exists only one nullary function, every operation is not also a function (I don't see how you'd deny this), or there doesn't exist a one-to-one correspondence between nullary operations and the elements of an arbitrary set. – Doug Spoonwood May 29 '11 at 01:59
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    @Doug: I did not claim "there is only one nullary function". You seem to think I did, but I did not. What I said is that there is only one function from the empty set to $A$. But, again, a nullary operation is not a function from the empty set to $A$. The domain of a nullary operation on $A$ is "the set of all functions from the empty set to $A$". It is this set which contains one and only one element. Don't confuse the set of all nullary operations with the set of all functions from the empty set to $A$. The former is $A^{(A^0)}$; the latter is $A^0$. – Arturo Magidin May 29 '11 at 02:02
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    @Doug: This is precisely the mistake I was trying to tell you about in my original comment. You are saying that "a nullary operation is a function from the empty set to $A$". It's not. Just like a binary operation is not a function from ${1,2}$ to $A$, it's a function from $A^2$ to $A$. – Arturo Magidin May 29 '11 at 02:05
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    @Doug: 23 hours passed since you asked the question until you thanked Arturo for the explanation of what a nullary function is. Do you still think it is a good idea of presenting groups using a definition that involves that notion? – Mariano Suárez-Álvarez May 29 '11 at 02:20
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    @Doug: I hope it's really the case that this is clear now. If not, then don't just say "thank you"; keep asking until I finally manage to make myself clear enough! – Arturo Magidin May 29 '11 at 02:26
  • @Mariano Well, perhaps not using a nullary function. However, I still think it a good idea to present groups using a unary function instead of an inverse of an element. As much as we can do so, in my opinion, I think it a good idea to either write proofs in a fully formal manner, or write proofs in such a manner that the formalization of a proof can easily get done. Proofs using a definition like those of definition 2 with only universal quantifiers, in my opinion, come as easier to formalize than those of definition 1, since existential elimination, in my opinion, comes as the hardest.. – Doug Spoonwood May 29 '11 at 02:33
  • rule to use, in say a natural deduction context, which I feel easier to use than any axiomatic system in writing a fully formal proof. If proofs either don't come as fully formal, or we can't see how they could get formalized, then I feel they merely convince. But, then, anything can qualify as a proof, and so long as someone convinces him/herself of something, there exist proofs of complete and utter rubbish. Additionally, if proofs in general just convince, then a proof consists of a psychological object and mathematics and logic fall prey to the dangers of psychologism. Though... – Doug Spoonwood May 29 '11 at 02:38
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    @Doug: Have you ever actually written down a proof in "fully formal manner"? Or, say, seen the fully formal proof that $1+1=2$ in Whitehead and Russell's Principia? They are far harder to follow and understand than what is considered to be a good, solid mathematical proof. I'm just going to have to disagree vehemently with your take on this, and point simply to just how completely you managed to confuse yourself (twice!) on the subject of nullary operations. – Arturo Magidin May 29 '11 at 02:40
  • formal proofs, and proofs which can get seen to get formalized easily, may seem more difficult at first, they both can get checked. Since they can get checked (how does one check a proof which just convinces? Consult the opinion of others? But a proof is not an appeal to opinions!), the dangers of psychologism get avoided. Also, I think knowing the logic behind a proof more important than knowing the result of a proof. Proofs which just convince emphasize, in my opinion, results too much over the reasoning involved. Formal proofs emphasize the reasoning. Though... – Doug Spoonwood May 29 '11 at 02:44
  • I will say that Arturo has a very good point about homomorphisms here, since checking does seem easier using definition 1 for homomorphisms. However, in the context of proofs for the algebra in general, not just between examples of the algebra, full formal proofs seem easier using a definition more like the second than the first in the original post. – Doug Spoonwood May 29 '11 at 02:48
  • @Arturo Yes, I have. See here http://spoonwood.xanga.com/747178518/a-simpler-proof-of-one-of-the-cancellation-laws-of-group-theory/. I don't think Principa a good example, since 1. uses infix notation which increases the number of symbols needed 2. uses axioms for logic, which significantly makes proving things in logic harder than say a non-axiomatic natural deduction style system does. Unless you have read exactly as many fully formal proofs as you have informal proofs, I don't see how you can adequately judge which comes as easier (this goes for me too, since I've read more informal.. – Doug Spoonwood May 29 '11 at 02:54
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    @Doug: I don't think we understand the same thing by either "proof" or "fully formal proof". As such, I will simply reiterate my intention to disagree vehemently with your take on the matter. – Arturo Magidin May 29 '11 at 02:57
  • than formal proofs). Though I may have come as confused about nullary operations, I haven't seen you present a fully formalized proof, rather than an informal one, concerning nullary opreations. Since informal proofs generally don't expose the principles of logic used in pretty much any way (the exact replacements used, the exact substitutions used in a proof), it comes as no surprise that confusion may result. So, confusion as a result of your informal reasoning here, doesn't come as a surprise here. However, if you presented a formal proof (I don't claim this is possible here) of – Doug Spoonwood May 29 '11 at 02:59
  • a statement in general, then I know that either I can verify everything in the proof myself (given the time needed to understand every step, of course), or I can reject your claim to actually having a proof. With informal proofs in general, I simply don't have a way to know if an error has creeped in somewhere and don't have a systematic way to reject a claim to having a proof. – Doug Spoonwood May 29 '11 at 03:05
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    @Doug: You were confused well before I entered the picture, as evidence by the incorrect statement in your original post, so kindly don't blame me or my "informal proofs" for that confusion. If you feel so strongly, I'll be happy to remove all the confusion I've introduced by deleting my participation and letting you be. – Arturo Magidin May 29 '11 at 03:05
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    @Doug: Also, I don't think the link you provide is for a valid proof of the statement you claim to be proving. If I read it correctly, what you proved is not left cancellation from the assumption of associativity, left inverses, and left identity. That statement would be (@x@y@z(Gxy=Gxz->y=z)). Instead, what you claim to have proven seems to be (@x@y@z Gxy=Gxz)->(@y@z y=z). This is not left cancellation, since (@x@y@z(p->q)) is not equivalent to (@x@y@zp)->(@x@y@zq). – Arturo Magidin May 29 '11 at 03:14
  • Though the last statement you claim holds, what you said before that does not. Since we have @x@y@z, Gxy=Gxz AND y=z, have x, y, and z bound, while "p" and "q" remain free in (@x@y@zp)->(@x@y@zq). Since the quantifiers in (@x@y@z(Gxy=Gxz->y=z)) bound over a conditional, they bound all instances of "x", "y", and "z" in both the antecedent and the consequent. So, (@x@y@z(Gxy=Gxz->y=z)) comes as logically equivalent to saying ((@x@y@z Gxy=Gxz)->(@y@z y=z)), as one can formally prove. As a simpler example, @p@qCKpqKqp in Polish notation logically comes as equivalent to C@p@qKpq@p@qKqp. – Doug Spoonwood May 29 '11 at 04:07
  • @Doug: Your proof has as a consequence that the group has only a single element (the statement @y@z(y=z) gives that all elements of the group are equal). You have proven a correct theorem, that states: "If $G$ is a group in which for every $x$, $y$, and $z$ we have $xy=xz$, then the group has a single element." Unfortunately, that statement is not the statement of left-cancellation. In short: what you think you've proven and what you've actually proven are not the same thing. – Arturo Magidin May 29 '11 at 17:40
  • @Doug: So, just to clarify: I'm not saying the proof is incorrect; I'm saying it's not a proof of left cancellation, it's a proof of something else. – Arturo Magidin May 29 '11 at 17:47
  • @Doug: And you are just incorrect to claim that (@x@y@z(Gxy=Gxz -> y=z)) is logically equivalent to (@x@y@x Gxy=Gxz)->(@y@z y=z). Sorry, but that's just wrong. In the semigroup with two elements $a\neq b$ in which the multiplication rule is given by $xy=x$ regardless of what $a$ and $b$ are, (@x@y@z Gxy=Gxz) is false, so (@x@y@z Gxy=Gxz) -> (@y@z y=z) is true. But @x@y@z(Gxy=Gxz -> y=z) is false, since the semigroup does not enjoy left cancellation. – Arturo Magidin May 29 '11 at 18:34
  • If the statements aren't equivalent, then how can I demonstrate what gets proven in this post: http://spoonwood.xanga.com/748810655/universal-quantifier-switch/ for an arbitrary predicate? Additionally, if not equivalent, then what do the quantifiers in (@x@y@z(Gxy=Gxz -> y=z)) bound over? I would hope you meant ((@x@y@x Gxy=Gxz)->(@y@z y=z)). Since we have xy=x, Gpq=p, Gpr=p. So, Gpq=Gpr is true. Since we had free variables, the hypothesis (@x@y@z Gxy=Gxz) follows as true. However, the conclusion comes as false, since left-cancellation fails for G:G11=1, G12=1, G21=2, G22=2. – Doug Spoonwood May 29 '11 at 23:34
  • @Doug: Sorry; I wrote something wrong, as you correctly point out here. In general, universals don't distribute over implications, and this has nothing to do with whether the variables are free or not. @x((x is even) -> (x is odd)) is false in the naturals, but (@x(x is even)) -> (@x(x is odd)) is true. I'll now point out that your "fully formal" proof was not understandable to me. In any case, I don't see much to do here except to disagree. – Arturo Magidin May 30 '11 at 01:17
  • @Doug: I think this is an example. Take your model as {1,2,3}. Define G11=G12=2, G13=3, and G2x=G3x=1 for all x. Since G11=G12 but 1=/=2, left cancellation does not hold, so @x@y@z(Gxy=Gxz->y=z) is false. However, @x@y@z(Gxy=Gxz) is also false, since G11=/=G13. And @y@z(y=z) is false. So (@x@y@z(Gxy=Gxz)) -> (@y@z(y=z)) is true, but @x@y@z((Gxy=Gxz)->(y=z)) is false. – Arturo Magidin May 30 '11 at 01:40
  • I think that works as a counterexample. But, what in the world do the quantifiers "@x", "@y", "@z" in @x@y@z((Gxy=Gxz)->(y=z)) bound if they don't bound all equiform instances of "x", "y", and "z" in the conditional? – Doug Spoonwood May 30 '11 at 02:54
  • @Doug: I don't understand what you are asking. The fact is, here we seem to have an instance in which @x@y@z((Gxy=Gxz)->(y=z)) is false, but (@x@y@z(Gxy=Gxz))->(@y@z(y=z)) is true. So the two statements are not logically equivalent, like I said originally. Whether or not your linked "universal quantifier switch" is correct, or whether or not it applies, I do not know because I don't plan to spend the time I would require to decypher it now. I don't see how what the quantifiers bound is relevant. It's clear to me that your asserted equivalence is false. – Arturo Magidin May 30 '11 at 02:57
  • @Doug: Never mind, I just tried. I think your "universal quantifier switch" is incorrect. @x@y(Pxy -> Pyx) is the statement that P is symmetric. (@x@yPxy)->(@x@y Pyx) is the statement that if P is total then P is total. Again, not equivalent. Take your model to be {1,2}, and Pxy to be true if and only if $x\lt y$. Then @x@y(Pxy->Pyx) is false, but (@x@yPxy)->(@x@yPyx) is true. And I think your error is that line 14 is incorrect. You have an assumption of the form "A implies B", but then you are taking A as an assumption. Try it with the example I just gave, I think you'll see the problem. – Arturo Magidin May 30 '11 at 03:03
  • I do see a problem with the example you gave me, thanks! But, an assumption of the form A implies B doesn't take A as an assumption, but only the conditional as the assumption. The antecedent of it could come as false. Also, Cpq, Cqr, p, q, r, Cpr, CCqrCpr, CCpqCCqrCpr takes two conditionals as assumptions, and qualifies as a proof in propositional calculus. Also, a proof like "Cpq, p, q, Cpq, CCpqCpq" has an assumption of the form Cpq, and then p taken as an assumption. Maybe my problem lies in thinking of predicate calculus too much like the propositional calculus with quantifiers. – Doug Spoonwood May 30 '11 at 04:20
  • Wait, in the proofs I gave I didn't specify a domain, so they don't necessarily hold for all domains. That there exist domains where they don't hold doesn't pose a problem necessarily, since just because a formula holds in a propositional calculus with domain {0, 1} that doesn't mean it has gotten claimed as holding in domain [0, 1]. The examples you provided happen outside of truth-operational domains, since your first G is not a truth-operation, nor is mapping {1, 2} to {T, F} a truth-operation. So, the last proof, I think, just got refuted as having that domain. – Doug Spoonwood May 30 '11 at 04:53
  • @Doug: I am well aware that taking an assumption of the form A->B is not the same as taking A as an assumption. I am pointing out that your "proof" of "universal quantifier switch" is false, because line 14 is unjustified. Line 14 asserts that @x@yPxy is an "assumption", but it is not at that stage of the proof. In any case, "universal quantifier swtich" is incorrect. @x@yPxy->@x@yPyx is a tautology. @x@y(Pxy->Pyx) is not a tautology; the two statements are not logically equivalent. – Arturo Magidin May 30 '11 at 12:31
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    @Doug: I have no further interest in this. So far, I've found errors in two of your asserted "fully formal proofs", despite your proclamation that it is only when you have "fully formal proofs" that you can be sure you are correct. Not only that, but your response is to insist that those proofs are correct, even in the face of simple counterexamples. You claimed you were proving that two statements are logically equivalent. That should mean they hold in all interpretations. Otherwise, your definition of "logically equivalent" is completely alien to mine. I will not reply any further. – Arturo Magidin May 30 '11 at 12:35
  • Assumptions don't need justification, and line 14 did qualify as an assumption, as it had scope to it. "Take your model to be {1,2}, and Pxy to be true if and only if x<y. Then @x@y(Pxy->Pyx) is false, but (@x@yPxy)->(@x@yPyx) is true." No, the last part isn't true, since P12=T, while P21=F, so ((@x@yPxy)->(@x@yPyx)) is false. So, that's not a counterexample. A proof of two statements as logically equivalent simply does not mean that such holds in all interpretations. At best, that one works for consistent interpretations. But, nothing prevents inconsistent ones. – Doug Spoonwood May 30 '11 at 13:39
  • @Doug: The interpretation is consistent, since it has a model. As to your first criticism, it's incorrect. @x@yPxy is false in that model, since P21 is false. @x@yPyx is false in that model, since P21 is false. Therefore, (@x@yPxy)->(@x@yPyx) is true, because it is an implication where the antecedent is false. So, yes, that is a counterexample. If two statements are logically equivalent, then they must both evaluate to "true" or both evaluate to "false" in any model, and I gave you a model, not an "inconsistent interpretation". @x@yPxy->@x@yPyx is a tautology; @x@y(Pxy->Pyx) is not. – Arturo Magidin May 31 '11 at 00:00
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    @Doug: But don't take my word for it. Post your proof of "universal quantifier switch" and I'll keep quiet. It has nothing to do with scope. But granted, I did not understand your argument (like what the bars on the left meant). The line I don't understand is 19. But it doesn't matter; the claim is false, as evidence by a single model where it fails. The "universal quantifier switch" is false, the two formulas are not logically equivalent; and your group theory proof is not a proof of the cancellation law. Period. – Arturo Magidin May 31 '11 at 00:05
  • @Doug: In the "@-out rule", are you allowed to substitute in constants? In the theory of groups, the following is a theorem: (@x(x=e))->(@y@z(y=z)), where e is the identity element constant. However, if we "@-out" the @x and replace x by e in the subformula, we get (e=e)->(@y@z(y=z)); this is not a theorem for groups. You can't eliminate universal quantifiers in subformulas, especially premises of implications, so haphazardly. In any case, I think you can prove to your own satisfaction that (@x@yPxy)->(@x@yPyx); that is, this implication is a tautology. Are you claiming @x@y(Pxy->Pyx) is also? – Arturo Magidin May 31 '11 at 01:35
  • @Arturo Line 19 in the "universal quantifier switch" argument might not work, I agree. Yes, in the @-out rule you can substitute constants. But, with the model you gave from @x@yCPxyPyx, I use-out twice @yCP1yPy1, and then CP12P21=Ctf=f, so I can deny @x@yCPxyPyx. That all said, the first counterexample you gave to the quantifier switch hypothesis worked. But, I've fixed the group theory proof without using a "quantifier switch", and put it in the comment section in this post: http://spoonwood.xanga.com/747178518/a-simpler-proof-of-one-of-the-cancellation-laws-of-group-theory/ – Doug Spoonwood Jun 01 '11 at 03:09
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    @Doug: At least the last line is now a correct statement of left cancellation; note that there is a typo at the very end: the last formula should be y=z, not x=z. I did not check the argument, though. As for the "universal quantifier switch", as I understand the universal instantiation rule, it only applies to formulas of the form @x(...), not to subformulas (though it can be done for some subformulas, e.g., when you have conjuctions). But since we are now in agreement that your original proof did not establish left cancellation, and that the u.q.s. is incorrect, I think we're done. – Arturo Magidin Jun 01 '11 at 03:19
  • @Arturo Thanks for pointing out the typo! – Doug Spoonwood Jun 01 '11 at 04:06
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Universal algebra presentations choose the algebra operations so that the general universal algebraic notions of homomorphism and substructure agree with the classical algebraic notions. Thus, for groups, to ensure that the universal notion of substructure yields a subgroup, we must include the inverse operation in the signature, else substructures would merely be subsemigroups (or submonoids if 1 is in the signature).

There is no need to do this in classical algebra textbooks because no attempt is made to give a universal definition of homomorphism or substructure. Rather, such definitions are presented in an ad-hoc manner - given specially for each type of algebraic structure defined (groups,rings, lattices, etc). It is only when we attempt to unify these concepts for various types of algebrac structures that we are forced to pay close attention to such matters.

Also worth emphasis: in universal algebra, we prefer equational axiomatizations, i.e. those given by purely $\rm\color{#c00}{universal}$ axioms (e.g. $\rm\color{#c00}{\forall\,x,y\!:}\ f(x,y) = g(x,y)),\,$ because they allow is to apply strong theoretical results about such classes of algebras (known as varieties or equational classes). For example, we can apply Birkhoff's completeness theorem, that an equation is true in all models iff it is provable by the usual rules of equational logic. Here is a concrete example. Jacobson published a model-theoretic proof of a theorem that any ring satisfying the identity $\rm\:x^n = x\:$ is commutative. By Birkhoff's completeness theorem there exists a purely equational proof of this theorem, but such a proof took a long time to be discovered, and has yet to be published (the case $\rm\:n = 3\:$ is a common moderately difficult exercise).

Bill Dubuque
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  • But, what if you want to actually write fully formal proofs of say elementary theorems (those capturable by first-order predicate calculus with equality) in such a text? Does a definition like the first really come as easier to fully formalize than the second? I think not, since the use of existential elimination say in a natural deduction style system, in my opinion, makes for a tricky business. – Doug Spoonwood May 29 '11 at 02:09
  • @Doug Indeed, universal algebra is primarily concerned with the rich theory if equational structures - see my edit above. – Bill Dubuque Jun 03 '11 at 03:07
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That style of definition is not used because one does not gain anything from using it. I would say that, in fact, it manages to obfuscate a very simple thing. For example, as you have probably noticed by now, the concept of nullary function is not exactly the clearest thing in the universe!

When one is studying universal algebra, or what not, well, in that context something is gained, but that is never done, as far as I know, in an algebra textbook...

Mariano Suárez-Álvarez
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  • I think formal proofs become simpler using definition 2, since you only need to concern yourself with universal quantifiers. In say the context of natural deduction, I think it simpler to use universal-introduction and universal-elimination than to use existential-elimination, although perhaps I just feel such easier. – Doug Spoonwood May 29 '11 at 02:06
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    @Doug: Have you looked into an actual textbook on group theory to see if anything would be simpler? Try rewriting, say, the proofs of Sylow's theorems, or the proof that the Frattini subgroup of a finite group isnilpotent using your definition 2... – Mariano Suárez-Álvarez May 29 '11 at 02:17
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Probably some principle of economy is involved.

Also, it is reasonable to develop the theory of groups and the theory of fields using roughly similar formalism. In axiomatizing fields, a multiplicative inverse operator would create difficulties.

One might note that in Group Theory, the usefulness for your $x'$ is quickly, if only implicitly, acknowledged. Almost any group theoretic work is littered with expressions of the shape $(x)^{-1}$.

I have not undertaken a survey, but it seems to me that what in logic one would call a constant symbol for the identity element occurs fairly often in the definition of group.

André Nicolas
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  • Well, do fields really qualify as an algebra, since you can't express the multiplicative inverse idea in terms of a sequence of existential quantifications, nor can you quantify over the entire reference set for a field? Maybe they don't. Maybe we should give up the idea that they do qualify as an algebra, and make them quasi-algebraic at best. Maybe not though. – Doug Spoonwood May 29 '11 at 02:15
  • @Doug Spoowood: I agree, they do not qualify. But I was pointing out that because classical algebra courses deal with, in particular, groups and fields, it gives reason not to treat inverse as a separate operator, since then the formal treatment of groups and fields would be different. – André Nicolas May 29 '11 at 02:20
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You could couch the existence of identity in terms of a 'nullary' operation, but this is probably obfuscatory. Quickly one notices this. Suppose 1 and 1' are both identities. Then

$$1 = 1*1' = 1'.$$

The element is unique. There is no need for an operation here.

ncmathsadist
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