How to prove multiplication of natural numbers by induction? Prove that there exists a unique function $g : \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ so that for all $x,y ∈ \mathbb{N}$
a) $g(x,1)=x$
b) $g(x,y')=x+g(x,y)$
So my attempt is(rough sketch):
Define $B$ to be the set of all $x \in \mathbb{N}$ so that one can find a set of function values $\{f(x,i):i \in \mathbb{N}\}$ so that for all $y$ both a) and b) holds( for the fixed $x$)
Base step: let $x=1$(keeping $x$ fixed), for every $y \in \mathbb{N}$, define $g(1,y)=y$. So $g(1,1)=1$ and therefore a) holds.
Now, $g(1,y')=1+g(1,y)=y+1$. So b) also holds.
Now let $x \in B$. To show: $x' \in B$. by definition $g(x',1)= x'=(g(x,1))'$
so a) holds.
Here is my problem, how do I show that $g(x',y')=x'+g(x',y)$
hint: Replace the notation $g(x,y)$ by $xy$ (multiplication of $x$ and $y$)
Thanks in advance.

There are few different ways you can define this type of “multivariate induction”. Since your version of natural numbers does not include $0,$ I will summarize a variation of mathematical induction on two variables accordingly as follows:


*

*$P(1,1)$

*$\forall x\quad P(x,1) \Rightarrow P(x+1,1)$

*$\forall x,y\quad P(x,y) \Rightarrow P(x,y+1)$


*

*$\forall x,y\quad P(x,y)$

It is readily verify that first two steps of this algorithm holds for your problem. Now for the last step assume that for some arbitrary $x,$ $g(xy)=xy.$ Then by $b)$ we have $$g(x, y+1)=x+g(xy)=x+xy=x(y+1)$$ as expected, and hence the result followed by mathematical induction.
