Let there be two functions, $f(x,y)$ and $g(x)$, such that $$f:\Bbb{R} \times \Bbb{R} \rightarrow \Bbb{R}$$ $$g:\Bbb{R} \rightarrow \Bbb{R}$$ Our initial set is $\{0\}$. If we define the functions like this $$f(x,y)=x \cdot y$$ $$g(x)=x+1$$ we can use them to generate $\Bbb{N_0}$ from $\{0\}$ (mathematical induction).
Now, lets create function $h(x)$ such that it maps $$\Bbb{N_0} \rightarrow S$$ where $S$ is set. Now, why is the function $h(x)$ defined like this invalid? $$h(0)=0$$ $$h(f(x,y))=f(h(x),h(y))$$ $$h(g(x))=h(x)+2$$ As you can see, I'm trying to use $h(x)$ (which is a recursive function) to generate $S$ from $\Bbb{N_0}$. I know that these rules are somehow contradicting each other, but I don't know how. That's my question: why are these rules contradicting each other?
This problem is taken from the book A Mathematical Introduction To Logic [2nd edition] (p. 38) by Herbert Enderton. This problem is not "copy-pasted" from the book, so I screenshotted it for those of you who want to see the original. You can see it here.
You need to show that the function $h$ you're constructing is in fact well-defined: I can't allow the rules I've set up to contradict each other.
For example, consider $h(6)$.
On the one hand, $6=f(2, 3)$ so we need to have $h(6)=h(2)h(3)$.
On the other hand, $6=g(5)$ so we need to have $h(6)=h(5)+2$.
On the other other hand, $6=f(1, 6)$ so we need to have $h(6)=h(1)h(6)$.
So in order for our rules to make sense, we'd better have $$h(5)+2=h(2)h(3)=h(1)h(6).$$ My point isn't that this is obviously false, but rather obviously worrying: we now need to prove, somehow, that we're never going to be able to apply the rules you've provided to produce contradictory results.
Towards actually showing that these rules lead to contradictory facts about $h$, it's a good idea to start by thinking about what you can say about the values of $h(0)$ and $h(1)$ ...