NOTE: THIS QUESTION IS NOT A DUPLICATE! THE ANSWER IS NOT IN THE RECOMMENDED QUESTION LINKED TO THIS POST
Reason: While the recommended question does mention $b_k(n)$, it never addressee why $b_k(n)\leq b_k(n-1)$, it instead glosses over this step and asks about a later step in the proof (why $b_k(n)=1$, a result, in which the answer to my question is necessary to conclude.)
I've been trying to prove the basis representation theorem:
$\forall k>1\forall n\in\mathbb{N}^+\exists!a\exists s(a:\downarrow s\to\underline{\downarrow}k\land a_s\neq 0\land n=\sum_{i=0}^sa_ik^i)$
Note: $\downarrow s=\{x\in\mathbb{N}|n\leq s\}$ and $\underline{\downarrow}k=\{x\in\mathbb{N}|n<s\}$
I've searched online and the same proof seems to be used everywhere for this.
Summary of my Work
First I assumed an arbitrary $n$ has a representation $a$ with domain $\downarrow s$
$$\exists s\exists a:\downarrow s\to\underline{\downarrow}k(a_s\neq 0\land n=\sum_{i=0}^sa_ik^i)$$
Next, since $\geq$ well-orders $N$, and $a_s\neq 0$, there must be a maximum number $t$ that doesn't map to $0$. $$\exists t\leq s(a_t\neq 0\land\forall x\leq t(a_x=0))$$
From there I split the sum : $$n=\sum_{i=t+1}^s a_ik^i+a_tk^t$$
I ended up with: $$n-1=\sum_{i=t+1}^sa_ik^i+(a_t-1)k^t+\sum_{i=0}^{t-1}(k-1)k^i$$
I next defined a new sequence $a'$ for $n-1$ so that:
$x<t\implies a'(x)=(k-1)$
$x=t\implies a'(x)=k$
$x>t\implies a'(x)=a(x)$
Thus ... $$n-1=\sum_{i=t+1}^sa'_ik^i+(a'_t)k^t+\sum_{i=0}^{t-1}a'_ik^i$$ $$n-1=\sum_{i=0}^sa'_ik^i$$
The following three cases complete the representation of $n-1$
$t<s\implies a'_s=a_s\neq 0$
$t=s\land a_t>1\implies a'_s=a_t-1\neq 0$
$t=s\land a_t=1\implies a'_{t-1}=k-1\neq 0$
So that any $n$ having a representation always implies $n-1$ has a representation: $$\exists s\exists a:\downarrow s\to\underline{\downarrow}k(a_s\neq 0\land n=\sum_{i=0}^sa_ik^i)$$ $$\implies$$ $$\exists s\exists a:\downarrow s\to\underline{\downarrow}k(a_s\neq 0\land n-1=\sum_{i=0}^sa_ik^i)$$
The Problem: Uniqueness
The next part of every proof I've seen on this theorem is that the above would imply that the number of representations of $n$ is always less than or equal to $n-1$ $$b_k(n)\leq b_k(n-1)$$
Without explanation this seems like a HUGE jump in logic.
Let's say there are two representations of $n$, let's call them $r_1n$ and $r_2n$.
How can we be sure that they lead to DIFFERENT representations of $n-1$?
If we can't be sure of that then how can we deny that:
$$b_k(n)>b_k(n-1)$$
?
