If Then Constraint Linear Programming
My solution requires some kind of hack in the model, a Big-$M$ value, which is pretty common, but also a tolerance $\varepsilon$, which is not very desirable. I hope that I do not miss something easy here but reckon there is no way around it.

To model a condition of the form $(X \wedge Y) \to Z$ for binary $X, Y, Z$, you can use $$\tfrac12(X + Y) \leq Z + \tfrac12.$$
We can verify this by inserting: $X=Y=1$ yields to $1$ on the left hand side, thus, $Z$ has to be at least $\frac12$, and since it is binary, it must be $1$. If either $X$ or $Y$ is $0$ (or both), we have at most $\frac12$ on the left hand side and in this case $Z$ might be either $0$ or $1$.

We also need to transform the constraint $B \leq m$ as a binary variable in order to use the template above. This can be done by introducing a new binary variable $B'$ which must assume the vaue $1$ iff $B \leq m$. We also need a Big-$M$ value which must be greater than any value that $B$ can assume (note that for numerical reasons in practice, $M$ should be not too large). Moreover, we have $\varepsilon$, a small tolerance value to get a strict $&gt;$-inequality (in practice you would choose the difference to the next greater number from $m$ that can be represented exactly on your machine). We now add the constraints $$B \leq m + (1-B')M\tag{1}$$ $$B \geq m + \varepsilon - B'M\tag{2}$$
$(1)$ guarantees that $B' = 0$ if $B &gt; m$, $(2)$ guarantees that $B' = 1$ if $B \leq m$. We verify this again: If $B \leq M$, $(1)$ and $(2)$ can be satisfied with $B' = 1$, but not with $0$. If $B &gt; m$, we need $B' = 0$ to satisfy $(1)$ and then $(2)$ is satisfied since $B$ is at least $\varepsilon$ larger than $m$.

Putting it all together, we obtain your constraint
$$
\tfrac12(A + B') \leq C + \tfrac12,\\
B \leq m + (1-B')M,\\
B \geq m + \varepsilon - B'M.
$$