Proof of recursive formula for "fusible numbers" The set of fusible numbers is a fantastic set of rational numbers defined by a simple rule. The story is well told here but I'll repeat the definitions. It's the formula on slide 17 that I'm trying to understand. 
Define $\displaystyle a \oplus b = \frac{a+b+1}{2}$. A number is fusible if it is $0$ or if it can be written as $a \oplus b$ where $a, b$ are fusible and $|a-b|<1$. Let $F$ be the set of fusible numbers. More formally, $F$ is the intersection of all sets of real numbers that are closed under $\oplus$ applied to arguments at a distance at most 1. 
The set $F$ is a well-ordered set of non-negative rational numbers. The proof that it's well-ordered isn't included in the PDF file I linked to, but it's not hard to show this. (It wouldn't be true if we hadn't insisted on the condition $|a-b|<1$, by the way.)
Amazingly, the order type of $F$ is $\varepsilon_0$. It's also true that $F$ is closed under ordinary addition; this isn't hard to prove either but I don't know if it plays a part in what follows.
Because $F$ is well-ordered, we may define $f(x)$ to be the least fusible number greater than $x$, for any real $x$. Further, set $m(x) = f(x)-x$. We obviously have $m(x) = -x$ for $x<0$, while for $x \geq 0$, it is posited that
$$m(x) = \frac{1}{2}m(x-m(x-1))$$ 
The question is: why is this last formula true?
I'm able to show one of the necessary inequalities, namely that $\displaystyle m(x) \leq \frac{1}{2}m(x-m(x-1))$: 
Given $x$, observe that 
$$(x-1+t) \oplus (x-t+u) = x + u/2$$ 
Take $t = m(x-1)$, which guarantees that ($x-1+t$) is indeed fusible. Now set $u = m(x-t)$ which makes ($x-t+u$) fusible as well, and the distance between those two fusible numbers can't be greater than $1$. It follows that ($x+u/2$) is fusible, and so $m(x)$ is at most $u/2$ for that particular $u$, which is indeed $m(x-m(x-1))$. 
The question, then, is: 
How can we prove that no other choice of $t$ yields an even smaller value for $m(x)$? 
It's not hard to show that there's no loss of generality in focusing on $x-1+t$ and $x-t+m(x-t)$, but greedily minimizing $t$ by setting $t=m(x-1)$ is not in any obvious way guaranteed to yield the minimum value for $m(x)$, as far as I can see. 
What am I missing?

It may be worth noting that a new paper on Fusible Numbers 'snuck onto' the arXiv earlier this year; it proves the order type $\epsilon_0$ and shows (unsurprisingly, given that) unprovability in PA of the statement 'there exists a next fusible number after $n$ for any $n\in\mathbb{N}$'. It doesn't entirely solve the minimal-gap question but establishes some new bounds, as well.  The paper is "Fusible numbers and Peano Arithmetic" from Jeff Erickson, Gabriel Nivasch, and Junyan Xu, at https://arxiv.org/abs/2003.14342 .
