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There is one thing that is used in a lot proofs but I have never heard of it as an axiom or a theorem. I'm talking about the fact that if $x$ and $y$ are numbers such that $$x=y$$ then $$f(x)=f(y).$$ And it seems true but I have never heard of it as an axiom (or theorem but I don't think that it can be a theorem). The only way (which I see) to prove it, is to somehow use the Peano axioms and one of ways to construct natural numbers but I don't imagine it in practice. So is there such an axiom or theorem?

Xander Henderson
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Юрій Ярош
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    that's part of the definition of a function. Specifically from a set theoretic standpoint, a function $f:A \rightarrow B$ is a subset $f \subset A \times B$ such that $\forall a \in A, \exists$ unique $ b \in B$ such that $(a,b) \in f$. We abbreviate $(a,b) \in f$ to $f(a) = b$. Hence by the uniqueness, if $x = y$ in $A$, we know that $(x, f(x))$ and $(y,f(y))$ must be the same ordered pair. Ergo, $f(x) = f(y)$ as members of $B$ – Joe Mar 19 '18 at 15:36
  • @Joe What definition you're talking about ? The one that uses tuples and sets or other ? – Юрій Ярош Mar 19 '18 at 15:38
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    @ЮрійЯрош That is the definition of a function from the point of view of set theory... – Xander Henderson Mar 19 '18 at 15:40
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    Related: Multiplying both sides of an equation in proofs. – Git Gud Mar 19 '18 at 15:41
  • @XanderHenderson I know. But it's, maybe, the most rigorous one. – Юрій Ярош Mar 19 '18 at 15:44
  • @GitGud Thanks. – Юрій Ярош Mar 19 '18 at 15:47
  • $x=y \implies f(x)=f(y)$ is trivially true by definition of function, if the function is defined for x=y, the reverse implication $ f(x)=f(y) \implies x=y$ is more interesting – user Mar 19 '18 at 15:48
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    @ЮрійЯрош What do you mean by "the most rigorous one?" It is the definition, from the point of view of set theory. How can it not be rigorous? – Xander Henderson Mar 19 '18 at 15:51
  • @XanderHenderson I meant that definitions like "it is a rule that assigns some input to some output" is not rigouros. – Юрій Ярош Mar 19 '18 at 15:53
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    Well, the logic I study has that as a logical axiom. So you can use it in any deduction and from soundness theorem it is valid as well. – Sorfosh Mar 19 '18 at 15:54
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    More formally $(x_1=y_1) \land (x_2=y_2)... \implies f(x_1.x_2...)=f(y_1,y_2...)$ – Sorfosh Mar 19 '18 at 15:56
  • @Sorfosh Thanks, actually, it was what I was interested in. Because I was thinking that something is left undefined. – Юрій Ярош Mar 19 '18 at 15:58
  • @ЮрійЯрош The definition you give is not nonrigorous; it is simply a bad definition. It can be made better by further defining what is meant by a "rule", an "input", and an "output," but those terms are left undefined in most texts that use such a definition. In the definition that Joe gives, all of the objects and terms used to define a function are also perfectly well-defined. – Xander Henderson Mar 19 '18 at 15:59
  • See First-order logic: equality axioms. – Mauro ALLEGRANZA Mar 19 '18 at 16:01
  • @XanderHenderson Yes. It seems to be rigorous but the only thing I'm not sure in the proof is words "the same ordered pair". – Юрій Ярош Mar 19 '18 at 16:06
  • @ЮрійЯрош From a set-theoretic standpoint, if you define the ordered pair $(x,y)$ to be the set ${x, {y } }$, being "the same ordered pair" follows easily as standard set equality. – Joe Mar 19 '18 at 16:16
  • @Joe Then it is clear for me, Thanks. – Юрій Ярош Mar 19 '18 at 16:19
  • @Joe Sorry, I have one more question. You said that "x=y in A so (x,f(x)) and (y,f(y)) are the same ordered pair". Do you use the fact that " x" and "y" are the same things there ? Or am I wrong ? – Юрій Ярош Mar 19 '18 at 16:38
  • @ЮрійЯрош Yes. Since $x$ and $y$ are sets, we have that ${x, {f(x)}} \in f$ and ${y, {f(y)}} \in f$. Using the fact that $x=y$ as sets, we have:${x, {f(x)}} \in f$ and ${x, {f(y)}} \in f$. But these must be the same set (uniqueness property), so ${f(x)} = {f(y)}$. Both of these sets have one element, so they must be the same element. Ergo, $f(x) = f(y)$ as sets. – Joe Mar 19 '18 at 18:54

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I think the answer to your question lies in the meaning of "equality". The statement $$ x = y $$ says (in everyday mathematics) that "$x$" and "$y$" are different names for the same object. So of course it follows that $$ f(x) = f(y) \ . $$

This doesn't depend on $x$ and $y$ being numbers, or even on $f$ being a function. It can be any expression in $x$ and $y$.

I say "in everyday mathematics" because if you're studying foundations you need to be much more careful about things and the names of things.

For an extended discussion, see this nice essay by Barry Mazur: http://www.math.harvard.edu/~mazur/preprints/when_is_one.pdf

Ethan Bolker
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  • "In everyday math" I agree with you. But I thought that there is something about it from the point of view of foundations. – Юрій Ярош Mar 19 '18 at 15:55
  • @ЮрійЯрош Perhaps. To find out here you'd have to make it explicitly a question about foundations - even more basic than Peano postulates. Ask it as a separate question rather than editing this one. You might enjoy the article I link to in my answer. – Ethan Bolker Mar 19 '18 at 16:08
  • OK, thanks for the article, I will read it. – Юрій Ярош Mar 19 '18 at 16:08