That is, $\forall x \ne 0 \in \mathbb{N},\exists y:y++ = x$.
I come up with this problem from a few sources that trying to prove the equivelance of WOP and Induction.
This one: https://www.math.wustl.edu/~chi/310notesIV.pdf. In part I it claim this, by a homework problem. However in my study experience, this lemma is proved using Induction.
This one:https://proofwiki.org/wiki/Equivalence_of_Well-Ordering_Principle_and_Induction. In part "WOP implies PFI", this line "So this minimal element of S′ has the form k+1 where k∈N." implicitly assume that this lemma is true.
I think we need to prove this lemma only using WOP without Induction, then we can safely say that WOP implies Induction.
Edit 1:
This idea comes up when I was told that WOP $\equiv$ Induction. Since Induction is the 5th Axiom of Natural Number, I think what we can use are only Axioms 1~4, then along with WOP as its replacement. Addition is not "invented" yet, what we can use is only successor notaton $++$.
Since WOP contains a prerequisite that there is an ordering(which is a pretty strong assumption) on Natural number set, I actually tend not to say that WOP $\equiv$ Induction. I also tend to think that this ordering need to be explictly constructed so that this ordering can be related to the actual numbers. Without an explicted construction, we do not have a way to link "the minimal element of this ordering" with the actual 0.
Edit2:
I spent this afternoon with one of my professor on this topic. We finally reach a conclusion that it is not true that Induction $\iff$ WOP. The reason is that WOP need a whole bunch of prerequisite structures before it can even make any sense, at least an ordering and addtion. Non of these can be achieved merely by WOP itself. However, Induction is much stronger; it is strong enough to build this ordering and all sort of structures by itself. In this sense, what we can say at most is that "along with a few other axioms to help building all these structures, the working scheme of WOP is equivelant to Induction."
I'm not sure of the detail but I think this comes down to the definition of "least". So far as I can tell $a \le b$ is defined by: there exist a natural number $c$ so that $a+c =b$.
But how is addition defined?
And so far as I can tell addition is defined recursively. $a+0 = a$ and $a + S(c) = S(a+c)$. And that's it.
That means that if $b = a+c$ and $c \ne 0$ then by definition $c= S(d)$ for some $d$ and that $b = S(a+d)$. So every number that can be written as sum with the second term non-zero, must have a predecessor.
Now.... Let $m$ be any non-zero natural number. Let $M = \{0, m\}$. By WOP that means we must have a least element. Let's assume that $m$ is the least element. So $m \le 0$ and there exists an natural $c$ so that $m + c =0$. If $c=0$ then $m =0$ but we assumed $m\ne 0$. So $0 = m+c; c\ne 0$ so $0$, by the paragraph above, must have a predecessor. But it is axiomatic that $0$ does not. So the least element is not $m$. so it must be $0$. So $0 \le m$ and there exist a $c$ so that $0 + c = m$. Again, $c$ can not be $0$ because $0 + 0 = 0\ne m$. So $m = 0+c;c \ne 0$ and, by the paragraph above, $m$ must have a predecessor.
So every non-zero natural number has a predecessor.
... I have to confess this feels like a rug pulled from beneath us. But... we must have an axiom that somehow says that the natural numbers are only those that can be created by iterations of successors from $0$. Defining a property recursively and then making an axiom that the natural numbers are only those that can have the property is equivalent. Induction seems less.... layered.... than WOP. But that's because it relies on defining addition.
==earlier answer===
Okay. Suppose there is a non zero natural number with no predecessor that means by WOP there is a least such $k$. $1,2,3$ all have predecessors so $k > 1,2,3$.
Now $k$, like all natural numbers, is larger than the natural numbers that are smaller than it. $k < m$ is, by definition, there exists a natural number $c\ne 0$ so that $m + c = k$. Let $M$ be $\{c\ne 0| \exists m\ne 0$ where $m$ has a predecessor and $m+c = k\}$
Now $M$ must have least element call it $d$. And that least element is not $0$. But if $m + d =k$ then $d + m =k$ and $d < k$ so $d$ has a predecessor. So let $e$ be the predecessor of $d$. Then $(m+1) + e=k$ But $e < d$ so $e\not \in M$. But we have $(m+1)+e = k$. The only way we can have $e\not \in M$ when there certainly does exist an $m+1$ so that $m+1 + e=k$ is if either $m +1 =0$. (That's impossible as no natural number has $0$ as a successor) or if $m+1$ doesn't have a predecessor (but $m+1$ has $m$ as a predecessor). Or if $e = 0$. But if $e = 0$ then $m+1 + 0 = k$ and $m+1 = k$. SO $m$ is a predecessor of $k$ contradicting that $k$ didn't have a predecessor.
Thus a contradiction. Thuse there is no natural non-zero number without a predecessor.