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I am just starting Apostol's Calculus Vol. 1, and I have no experience with rigorous mathematics. In his introduction, the first proof he gives is that if $a+b=a+c$, then $b=c$.

It says, "Given $a+b=a+c$. By Axiom 5, there is a real number $y$ such that $y+a=0$. Since sums are uniquely determined, we have $y+(a+b)=y+(a+c)$..." This part confuses me because it appears that he is essentially using substitution, but substitution hasn't been mentioned at all.

Before the theorem he is proving, he says it is assumed sums are uniquely determined and gives only these six axioms for the real-number system: commutative laws, associative laws, distributive law, existence of identity elements, and existence of reciprocals.

I'm not sure how the allowance of substitution follows from any of these. I mean, he explicitly states it follows from the assumption that sums are uniquely determined, but I don't follow how that allows substitution.

Edit: Apparently it's a property of equality called the substitution property which states "For any quantities a and b and any expression F(x), if a = b, then F(a) = F(b) (if both sides make sense, i.e. are well-formed)." https://en.wikipedia.org/wiki/Equality_%28mathematics%29

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  • $\begingroup$ All he is saying is first that every real number has an additive inverse. Here set $z = a+b$ then there is a $y$ with $y+ z = y+(a+b)=0$. By assumption $z$ is also $a+c$ so for this same $y$ we have $y + (a+c) = 0$. Now they are both equal to zero so $y+(a+b) = y+(a+c)$. $\endgroup$ Commented May 24, 2016 at 2:43
  • $\begingroup$ In proving substitution, isn't that using circular reasoning by saying $z$ and $a+b$ have the same inverse? Anyhow, after googling for a bit, on Wikipedia's page for equality it mentions something about a substitution property which states "For any quantities a and b and any expression F(x), if a = b, then F(a) = F(b)". Does this make it an axiom? $\endgroup$
    – user341840
    Commented May 24, 2016 at 2:50
  • $\begingroup$ What do you mean by substitution? He is just adding $y$ to both sides. $\endgroup$
    – user333870
    Commented May 24, 2016 at 4:08
  • $\begingroup$ Substitution is the definition of equality, which is defined not only for the real numbers. $\endgroup$ Commented Nov 27, 2018 at 15:39

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He's not using substitution, he's just adding $y$ to both sides. $y$ is the inverse of $a$. There's no substitution at all.

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  • $\begingroup$ Adding $y$ to both sides is equivalent to substituting $(a+b)$ with $(a+c)$ in the expression $y+(a+b)$. Besides, which axiom/assumption that he listed justifies adding $y$ to both sides? I googled a bit, and I think I found what I was looking for. Apparently there's a property called "the substitution property of equality" that allows this. en.wikipedia.org/wiki/Equality_%28mathematics%29 $\endgroup$
    – user341840
    Commented May 24, 2016 at 6:39
  • $\begingroup$ Adding $y$ to a real number is a function. The definition of a function says that if $x = z$, $f(x) = f(z)$. That is what allows you to add $y$ to both sides and preserve equality. $\endgroup$
    – Vik78
    Commented May 24, 2016 at 7:10
  • $\begingroup$ How would you prove "adding $y$ is a function" without using substitution, then? I have tried... Sps $x_1 = x_2$. But then to show $x_1 + y = x_2 + y$ we are using substitution. And we cannot say "but adding is a function" either, because that's what we are trying to prove. If we really had to use substitution, I would like to see a proof of the substitution property... To clarify: either a proof that the above is a function w/o substitution, or a proof of substitution itself. $\endgroup$ Commented Jun 12, 2021 at 3:48

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