Please ponder on an equilibrium of a very weak base, say $\ce{MOH}$. This base undergoes a very low ionization, even in its pure form (pure form implies that there isn't the presence of a solvent). The concentration of $\ce{OH-}$ ions, that our base generates, can be assumed to be $ \le 10^{-6} $. Now, let us add some water to $\ce{MOH}$ so as to dilute it. Ostwald's dilution law states that if the degree of dissociation ($ \alpha $) is very less as compared to unity, which in this case, is, it can be calculated by the equation $ \alpha = \sqrt{\frac{K_a}{C}} $. This leads to the conclusion that as I make the solution less concentrated (by dilution), $ \alpha $ increases.
But if I think of it in a different way, I get confused. For instance, please consider the simultaneous equilibria: $$ \ce{MOH <=> M+ + OH-} \tag{1}$$ $$ \ce{H_{2}O <=> H+ + OH-} \tag{2}$$ Since the concentration of $ \ce{OH-} $ ions our base generated in its pure form was comparable to the concentration of $ \ce{OH-} $ ions generated by water, they may cause the first equilibrium to be shifted in the backward direction, thus leading to a decrease in $ \alpha $ upon dilution. Is there any point I'm missing which makes me conclude the opposite of what one concludes from Ostwald's dilution law?