Naive Set Theory - proof of commutativity of products

Question

I am working through Halmos's Naive Set Theory on my own and trying to do all the exercises, including what are merely suggestions in the text. I am right now in section 13, which shows a derivation of addition, multiplication and powers from the set omega of natural numbers. Author states that 'Multiplication is associative and commutative. The proofs are straightforward adaptations of the ones that worked for addition'. I have tried to do them, and have managed to show associativity and the distributive property as well, but am stuck with commutativity of products. I have been working with this for 3 days with no success. Let me show what I have reached.

Commutativity of products

To show $ab=ba$ using induction, we try induction on $a$. Let $a=0$. By the way Halmos has defined products, it follows $m(0)=0$, so $b(0)=0$, but we get no direct indication of what $0b$ may be. For this we need to develop an intermediate step, showing by induction on $b$ that $0b=0$. Base case: $b=0$. Then we have $0\cdot 0= 0(0)=0$. Assume true for $k$, show it follows for $k'$: $0k'= 0k+0 = 0+0 = 0$.

So now we can make the base case of $ab=ba$, induction on $a$, $a=0$. $0b=0$, as shown above, and $b(0)=0$.

For going to the induction step, we assume true for $k$ ($kb=bk$), show it is true for $k'$ ($k'b=bk'$). Halmos's definition of multiplication only allows you to advance when the second element of the product is a successor, so $bk'= bk+b = b+bk$ (I've shown commutativity of sums, so this is valid) $= b+kb$ (Induction Hypothesis). I can't see how to simplify this further, so my hunch is the next step would be to try to show through some intermediate step that $k'b = b+bk$, with induction on $b$. I've tried this for hours and it doesn't seem to work out. I would be grateful for any indirect tips and suggestions on how to proceed, or information that i am following a dead end.

Some comments state that this has already been answered, but in the questions they link, I don't think the scenario is exactly the same. In previous chapters, Halmos has stated 5 Peano axioms, none of which state details or properties of 1 (although using the machinery, I guess I could show that $0'=1$ and try to employ it in the proof). $1$ is mentioned later -haven't gotten to that yet- for powers, but I didn't expect to need to use it for proofs that are 'straightforward adaptations' of the ones for addition, which didn't use it.

Answer 1

Fix $a$. We want to prove that $ab=ba$ by induction on $b$. The base is done. As in the case of addition, we need to show a preliminary result. Halmos notes that to prove commutativity of addition, we need to show that $0+n=n$ and that $m^++n=(m+n)^+$. The analogs of these for multiplication are to show that $0x=0$ (which you have done); and that $y^+x = yx+x$.

Lemma. Fix $y$. For all $x$, $y^+x = yx + x$.

Proof. We have that $y^+0 = 0$; on the other hand $y0+0 = 0+0=0$. So the equation holds for $x=0$.

Assume that $y^+x = yx+x$. We prove that $y^+x^+ = yx^+ + x^+$. We have $$\begin{align*} y^+x^+ &= (y^+x)+y^+ &\text{(definition of multiplication)}\\ &= (yx+x)+y^+ &\text{(induction hypothesis)}\\ &= yx + (x+y^+) &\text{(associativity of addition)}\\ &= yx + (x+y)^+ &\text{(definition of addition)}\\ &= yx + (y+x)^+ &\text{(commutativity of addition)}\\ &= yx + (y+x^+) & \text{(definition of addition)}\\ &= (yx+y) + x^+ &\text{(associativity of addition)}\\ &= (yx^+) + x^+ &\text{(definition of multiplication)}\\ \end{align*}$$ as desired. $\Box$

Theorem. Fix $a\in\omega$. Then for all $b\in \omega$, $ab=ba$.

Proof. Induction on $b$. We have $a0=0 = 0a$, establishing the base.

Assume that $ab=ba$. Then $$\begin{align*} b^+a &= ba+a &\text{(by Lemma)}\\ &= ab+a &\text{(by the induction hypothesis)}\\ &= ab^+ &\text{(by the definition of multiplication)} \end{align*}$$ which completes the induction. $\Box$