# Is my Peano Axiom proof correct?

I have recently begun reading Terry Tao's Real Analysis 1. The Peano Axiom proofs in the book are very new to me. Because of this, I have little intuition as to whether my proofs are correct. In addition to this, solutions to the exercises are not available in the text and I don't know anyone who is well-versed in this. This is the motivation behind my question. Any feedback is welcome! Thank you very much for taking the time, I truly appreciate it.

Here are the axioms as stated in his book:

Axiom 2.1: $$0$$ is a natural number.

Axiom 2.2: If $$n$$ is a natural number, then $$n++$$ is also a natural number.

Axiom 2.3: $$0$$ is not the successor of any natural number; i.e., we have $$n++\not= 0$$ for every natural number $$n.$$

Axiom 2.4: Different natural numbers must have different successors; i.e., if $$n,m$$ are natural numbers and $$n\not=m$$, then $$n++\not=m++$$. Equivalently, if $$n++ =m++$$, then we must have $$n=m.$$

Axiom 2.5: Let $$P(n)$$ be any property pertaining to a natural number $$n$$. Suppose that $$P(0)$$ is true,and suppose that whenever $$P(n)$$ is true, $$P(n++)$$ is also true. Then $$P(n)$$ is true for every natural number $$n.$$

We can also assume that recursive definitions are well defined.

Definition of addition: Let $$m$$ be a natural number. To add zero to $$m$$, we define $$0 +m:=m$$. Now suppose inductively that we have defined how to add $$n$$ to $$m$$. Then we can add $$n++$$ to $$m$$ by defining $$(n++ ) +m:= (n+m)++.$$

Additionally, the following have been proven:

Lemma 2.2.2: $$n+0=n$$ for all natural numbers $$n$$.

Lemma 2.2.3: For any natural numbers $$n$$ and $$m, n+(m++ ) = (n+m)++.$$

Proposition 2.2.4: $$n+m=m+n$$ for all natural numbers $$n,m.$$

Given this, is the following proof right?

Theorem 1: For any natural numbers $$a,b,c$$ we have $$(a+b)+c=a+(b+c).$$

My proof: We induct on $$c.$$ First, we verify the base case: $$(a+b)+0$$ The above equals $$a+b$$ by lemma 2.2.2. Additionally, this can be written as $$a+(b+0)$$ by lemma 2.2.2. This concludes the base case. Next, assume that there exists a natural number $$c$$ such that $$(a+b)+c=a+(b+c)$$ for all natural numbers $$a,b.$$ Then, for $$c++$$ we have $$(a+b)+(c++)=((a+b)+c)++$$ by lemma 2.2.3. By the inductive hypothesis, this becomes $$(a+(b+c))++.$$ Using lemma 2.2.3, this becomes $$a+(b+c)++.$$ Applying lemma 2.2.3 once more yields $$a+(b+(c++))$$ as desired.

$$\square$$

2. Instead of "assume that there exists a natural number $$c$$ such that...", you should write "take any natural number $$c$$, and assume that for all $$a,b$$, $$(a + b) + c = a + (b + c)$$". It is under these assumptions that you are going to show that the statement "for all $$a,b$$, $$(a + b) + c\!+\!\!+ = a + (b + c\!+\!\!+)$$" holds for the same $$c$$.
edit: One more comment about style. Instead of the phrase "this becomes", you can always write chains of equalities. For example, you can write "by Lemma 2.2.3, $$(a + (b + c))\!+\!\!+ = a + (b + c)\!+\!\!+$$". This way the reader doesn't have to guess which side/subexpression of the original expression you mean, which may improve readability in more complicated arguments.