1

I am confused about something that seems to be known as the "substitution property," which I take to be central to mathematical reasoning. Informally:

$$ \text{If } b = c \text{ then } a \oplus b = a \oplus c. $$

I don't remember ever seeing the above inference defined or proved, which leads to cognitive dissonance when trying to follow algebraic proofs that employ it. Where does it come from? If it depends on choice of foundations, what are the alternatives? Do I need to take courses in Mathematical Logic and Type Theory to understand what is going on here?

To capture the way that I currently reason about things, I have restated the problem below. I think that my confusion would be resolved if I could prove the following proposition:

Let $X$ be a mathematical structure of unspecified foundations.

For any symbol $a$, "$a$ is an $X$" is either true or false.

We will write "$a, b, \ldots$ are $X$" to mean $a$ is an $X$ and $b$ is an $X$ and $\ldots$

Let $\oplus_X$ denote a symbol that combines any two symbols $a$, $b$ such that:

  • if $a$, $b$ are $X$ then $a \oplus_X b$ is an $X$.

Let $=_X$ denote a symbol that combines any two symbols $a, b$ such that:

  • if $a, b$ are $X$ then $a =_X b$ is either true or false
  • if $a$ is an $X$ then $a =_X a$
  • if $a, b$ are $X$ then (if $a =_X b$ then $b =_X a$)
  • if $a, b, c$ are $X$ then (if $a =_X b$ and $b =_X c$ then $a =_X c$)

Proposition: If $a, b, c$ are $X$ and $b =_X c$ then $a \oplus_X b =_X a \oplus_X c$

[Note: fixed serious typo, had written If $a, b, c$ are $X$ and $b =_X c$ then $a \oplus_X b = a \oplus_X c$]

Is the proposition provable from the given definitions? If not, what are the missing ingredients? Are they a property of the chosen logic? Or of the choice of foundations? (e.g. Set theory, Type theory) If so, how? Or does the proposition always depend on the structure X? i.e. should the proposition always be proved?

My reading so far has lead me to the following unresolved ideas:

  • In high-school algebra, it would appear that the "substitution property" is taken as axomatic. Is this also the case in abstract Algebra? or in mathematical reasoning more generally?

  • Often, $X$ is defined as a Set with some additional structure (e.g. a Group), so perhaps some basic property of Sets is assumed?

  • In contexts involving equivalence classes, we must prove that the "operation" $\oplus_X$ is "well-defined". (One of my teachers used to say that really we are asking whether $\oplus_X$ is "defined" at all!) Perhaps this means that the "substitution property" is an implied property, or part of the definition of an "operation"? Although I found no such information in the definition of operation[1].

  • An unanswered question[2] mentions the "Axiom of Substitution" but I don't remember seeing such an axiom and I could not find a definition of it.

My background: I am a student of mathematics at the level of a not strong mid-to-upper level undergraduate. I also have an informal background in computer programming. I have not studied mathematical foundations or mathematical logic. I read Halmos' Naive Set Theory a long time ago, I have not read Bourbaki (but I can if you think it will help), I have heard of HoTT, I have seen a definition of natural deduction, and I know that sometimes "=" is used to mean canonical isomorphism. I would be comfortable if your answer involved these and other related concepts.

[1] https://en.wikipedia.org/wiki/Operation_(mathematics)#Definition

[2] Question about the axioms of equality, particularly about the axiom of substitution

Ross Bencina
  • 151
  • In most cases this is just the same thing as saying the function given by your operation is well-defined, which is baked into being a function. If some function $f$ is given by the rule $f(z) = a \oplus z$ and $b = c$ then, by well-definedness and the definition of $f$, we have $a \oplus b = f(b) = f(c) = a \oplus c$. Proving well-definedness is fairly common in math, and is often done up to some other notion of equivalence relation other than equality. In the most general case, you can read up on substitution in logic (there's a Wikipedia page on it, though I'm not sure if it's well written) – Brevan Ellefsen Nov 24 '22 at 23:42
  • As such, your proposition is basically just claiming $\oplus_X$ is well-defined on the equivalence relation $X$. Note this isn't always the case: as a trivial but important example, suppose we are working with disconnected surfaces and want to define a wedge sum. It is no longer defined, since for example we could take two copies of $S^2 \sqcup T^2$ (sphere and torus) and apply an isomorphism to flip the sphere and torus in space. If we then wedge sum a cylinder to the same place in space for both we get different things, so the wedge sum is only well-defined for connected surfaces. – Brevan Ellefsen Nov 24 '22 at 23:53
  • I think that you are overthinking this. Equality means equality in every way. It is a much stronger relation than mere equivalence. – Somos Nov 25 '22 at 00:03
  • 1
    Closely related: https://math.stackexchange.com/questions/2882730/is-there-a-proof-that-performing-an-operation-on-both-sides-of-an-equation-prese – Daniel Schepler Nov 25 '22 at 00:55
  • @Somos: could you explain how your comment remains true when we define the real numbers as equivalence classes of rational Cauchy sequences? as in https://proofwiki.org/wiki/Definition:Real_Number/Cauchy_Sequences – Ross Bencina Nov 25 '22 at 04:01
  • Please explain why it matters how real numbers are defined. – Somos Nov 25 '22 at 04:10
  • @Somos: if we define the reals axiomatically as a complete ordered field, then your statement "equality means equality" might be construed as true, since the equality you speak of is e.g. the first-order logic one in MarkSaving's answer. If on the other hand, we define the reals as equivalence classes of rational Cauchy sequences, two real numbers are equal iff they are representatives of the same equivalence class i.e. equality of reals is precisely the defined equivalence relation. – Ross Bencina Nov 25 '22 at 04:15
  • Does this answer your question? Is there a law that you can add or multiply to both sides of an equation? – MJD Nov 25 '22 at 04:19
  • 1
    @MJD: That was helpful, and answers my question to some extent, but the answers align very much to ZFC orthodoxy and don't address how things vary (or not) with different choices of foundations. – Ross Bencina Nov 25 '22 at 04:54

1 Answers1

5

Is the proposition provable from the given definitions?

Clearly not. If we’re working in set theory, all these axioms require is that $=_X$ is a (decidable) equivalence relation on $X$. There is no reason to think that an operation should respect all possible equivalence relations.

In first-order logic with equality, we have the following rule. We can derive $P(y)$ from the premises $x = y$ and $P(x)$. This is a basic rule/axiom of first-order logic with equality.

We can also always derive $x = x$. This is the reflexive property of equality.

In particular, let $P(x)$ be the proposition $a \oplus b = a \oplus x$. Then $P(b)$ is an instance of the reflexive property of equality, so we know $P(b)$. If we then assume $b = c$, it follows from the substitution property that $P(c)$ also holds. That is, $a \oplus b = a \oplus c$. So we have proved that if $b = c$, then $a \oplus b = a \oplus c$.

Type theories vary somewhat in their treatment of equality, but they usually have a version of the substitution property and reflexive property as axioms. Thus, the proof in type theory should look similar to the proof in first-order logic.

Mark Saving
  • 31,855
  • Thank you Mark, that is a very clear answer. However it makes me realise that I have been confused in not clearly distinguishing alternate frameworks: e.g. when working with the real numbers we may work axiomatically, with variables and equality defined by first order logic as you describe, or we may work with the definition of the reals as equivalence classes of Cauchy sequences, in which case equality of reals is defined as being members of the same equivalence class. In each case the equals sign means something different but in some sense we are talking about the same objects. – Ross Bencina Nov 25 '22 at 03:48
  • 1
    @RossBencina When we are discussing the reals defined using Cauchy sequences, equality has its usual meaning. Everything I’ve said applies to this case. You are confusing the real numbers with Cauchy sequences of rationals. A real number is not a Cauchy sequence; it is an equivalence class of Cauchy sequences. So it is meaningless to say “equality of reals is defined as being members of the same equivalence class”. Real numbers are the equivalence classes you speak of; they are not members thereof. – Mark Saving Nov 25 '22 at 06:52
  • apologies, I have not been precise. I understand that a representative of an equivalence class is not the equivalence class itself. My point was, when dealing with a object defined in terms of equivalence classes (e.g. the reals), we will at times be wrangling specific representatives, either to make use of the specifics of a representative, or to show that something is independent of the choice of representative. The example I have in mind is that Newton's Method applied to $x^2 - 2$ gives us a representative of the class usually called $\sqrt{2}$. – Ross Bencina Nov 25 '22 at 08:44
  • in any case, taking the reals as equivalence classes (i.e. sets of Cauchy sequences), surely the equality that is used to compare reals is the equality defined for Sets, not the first-order syntactic equality that you have so clearly described above. To be clear: by equality on sets I mean: two sets are equal if they are subsets of each other. Furthermore, establishing the equality of those equivalence classes, in the end, hinges on the equivalence relation that relates two Cauchy sequences as representing the same real number. – Ross Bencina Nov 25 '22 at 08:50