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The imported dyadic theorem says that for $u,v>0$ with $u/v$ irrational, every sufficiently large integer can be written as

$$\sum_{r\in R}\lfloor u2^r\rfloor+ \sum_{s\in S}\lfloor v2^s\rfloor,$$

where $R,S$ are finite subsets of the nonnegative integers. An index is used once in each ray; equal values at different indices are permitted by that theorem. Its recorded acceptance is on the dyadic claim page.

Fix $m,t,L,F$ as in L33 and put $\gamma=2^{1/m}$. A rational $m$th root of an integer is an integer, and no integer has $m$th power $2$ for $m\ge2$. Thus $\gamma$ is irrational and $1<\gamma<2$.

Choose an integer $j\ge L$ so large that

$$t\gamma^n(\gamma-1)>1\quad(n\ge j), \qquad \lfloor t\gamma^j\rfloor>\max(F\cup\{0\}).$$

These inequalities are possible because $t>0$ and $\gamma>1$. The floor values with indices at least $j$ are positive, strictly increasing, and outside $F$.

Set $u=t\gamma^j$ and $v=t\gamma^{j+1}$. Both are positive and $u/v=\gamma^{-1}$ is irrational. Since $\gamma^m=2$,

$$\lfloor u2^r\rfloor=\lfloor t\gamma^{j+mr}\rfloor, \qquad \lfloor v2^s\rfloor=\lfloor t\gamma^{j+1+ms}\rfloor.$$

The classes $j+mr$ and $j+1+ms$ are disjoint for $m\ge2$. Consequently each representation given by the dyadic theorem uses distinct indices of the original tail. Strict increase on that tail makes their integer values distinct as well. All those values are positive and avoid $F$, which proves L33.

For Problem 354(ii), choose $m=2$ and $t=\alpha$. Completeness of this single ray implies completeness after adjoining the $\beta$ ray for every $\beta>0$, whether or not $\alpha/\beta$ is irrational. The argument works with first index $0$ or $1$ and with either indexed or distinct-value sums. The threshold is allowed to depend on every fixed parameter and the finite deletion. No assertion for arbitrary $\gamma\in(1,2)$ follows.