Does "normal" induction work in any well-ordered set? This is a follow up from this question.
I want to prove whether the usual "induction theorem" works in a general  well-ordered set $A$:

If $(A,\preccurlyeq)$ is a well-ordered chain, $\Phi(\min A)$ is true, and for all $x\in A$, $\Phi(x)\Rightarrow \Phi(x^+)$,  then $\Phi(a)$ is true for all $a\in A$.

For $\mathbb N$ it works because apart from the successor, you can also evaluate the "precursor"
$$x^-=\max\{a\in A : \text{$a\preccurlyeq x$ and $a\neq x$}\}.$$
Indeed, one can proceed by contradiction, noting that if it is false, $\min\{a\in A : \lnot\Phi(a)\}$ exists (call it $a_0$), but then you get that $\Phi(a_0^-)$ is true, so $\Phi((a_0^-)^+) = \Phi(a_0)$ is true, a contradiction.
(In the case of $\mathbb N$, the maximum always exists because the corresponding set is bounded above and finite, hence my original question.)
When I see this stated for the general case, it is always phrased in the "strong induction" way, and the proof depends on that fact. Here is an example:

So is my (normal) induction statement true? I suspect not now that I think about it, since if we have order type $\omega+1$ (say) we never establish $\Phi(x)$ for the last element by chaining implications together as induction usually does for $\mathbb N$ (i.e., $\Phi(1) \Rightarrow\Phi(2)\Rightarrow\cdots$).
 A: No, your "normal induction statement" is not true.
Consider the following well ordered set $A$, $\{0,1\}\times\omega$, with the lexicographic order. That is, $(a,b)\lt (c,d)$ if and only if $a\lt c$, or $a=c$ and $b\lt d$. Think of it as "green" and "yellow" natural numbers, with all the green natural numbers going first, and all yellow natural numbers going later. The successor funtion in this set is $(a,b)^+ = (a,b^+)$.
This is well ordered: given any nonempty subset $X$ of $A$. If $X\cap \{0\}\times\omega$ is nonempty, then the preimage of the least element of the projection of this subset onto $\omega$ is the least element; otherwise, the least element of $X$ is the smallest "yellow" natural number that it contains. (That is, if $X$ has green natural numbers in it, then its least element is the smallest green natural number in it; otherwise, take the smallest yellow number in it).
Now consider the property $P(a,n)$ to mean "$a=0$" (i.e., "$n$ is green"). Clearly, $P(a,n)$ does not hold for all elements of $A$.
However: $P(0,0)$ holds. And if $P(a,b)$ holds, then $P( (a,b)^+) = P(a,b^+)$ holds, since $a=0$. Thus, your "normal induction statement" holds for $P$ in $A$. Yet $\forall x(x\in A\implies P(x))$ is false.

Note that your statement for "strong induction" can be simplified. In fact, you only need to prove that
$$\forall a\in A\Bigl( \bigl(\forall a' (a'\lt a\implies P(a'))\bigr)\implies P(a)\Bigr)$$
and that will prove $P(a)$ for all $a\in A$, since for $\min(A)$, the premise $\forall a' (a'\lt \min(A)\implies P(a'))$ is true by vacuity. So you don't need to separate out the least element of $A$, and then explicitly exclude it from the inductive step. See comments here and here. It's just that one usually needs to do "special cases" for the least element, but sometimes for other values instead.
