Now I have to heavily emphasize the fact that I have never studied differential algebra or the concept of other types of differentiation (which is what I believe is the concept behind a differential algebra). So, if I am abusing the terminology a little bit, please forgive me.

Let us define a differential algebra known as implied differentiation. It actually does not have a unique value. Let us denote the implied derivative operator as $I(f)$, where $f$ is any function being implied differentiation. This is of course, nonstandard terminology.

We define the solution set of $I(f)(x)$ to be $I(f)(x) := \{y| y = \lim_{h\to 0^+} \frac{f(x+h)-f(x)}{h} \lor y = \lim_{h\to 0^-} \frac{f(x+h)-f(x)}{h}\}$.

By this I mean that the evaluation of $I(f)$ at $x$ is any number in that set. Hence, the implied derivative is a multi-valued operator. It can return any number of functions depending on the continuity of $f$.

The general crux of this question is that I wish to determine whether or not the following statement is true. I'm pretty sure it is, though that's just intuition.

Is a function a solution to an ordinary differential equation if and only if it is a continuous solution to the associated implied differential equation?

By corresponding equations, I just mean that they are corresponding if they are the same except with all of the derivative operators replaced with the implied derivative operator. So, the equation $D(y) = e^x$ has a corresponding equation of $I(y) = e^x$.

I do not know for sure whether or not anyone will actually be able to prove this statement. I think it is a bit tricky, but even just some advice on how to approach this or how to begin would be greatly appreciated. It isn't for homework or anything like that. It's just a statement and concept I've developed in my head over the years and I want to determine its truthfulness.

**Note: if something equivalent to this or very similar that just makes this a special case has been proven in the past feel free to use that as an answer. I'm going under the (possibly mistaken) impression that this hasn't actually been proven before.**

*My attempts:*

I believe that a possible route to proving this statement might come by proving the following propositions.

**The solution sets of sub-differential equations are a super set of the solution sets of the corresponding differential equations**

**The solution sets of sub-differential equations are a super set of the solution sets of the corresponding implied differential equations**

**If a solution to a sub-differential equation is continuous then it is a solution to the corresponding differential equation**

**If a solution to a sub-differential equation is continuous then it is a solution to the corresponding implied differential equation**

I believe that the the first two propositions might follow trivially from the definition of the sub-derivative. The third one might have been proven in the past. I am unsure. The fourth one would then be the true meat and bones of the proof.

The subderivative is defined here: https://en.wikipedia.org/wiki/Subderivative

*Why this is important:*

The implied differential operator is defined via the limits; however, the purpose of that is to emulate a differential algebra wherein all step functions have a derivative of 0 and all the other normal rules are preserved. If it wasn't apparent, because of this any implied differential equation involving step functions is trivial to solve (at least in terms of the step functions themselves providing difficulty). If the statement were true, it would provide a new avenue by which to attack differential equations, some of which might be made trivial to solve via this method.