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I have the following lemma to show the derivative of f at x is D. 

lemma lm1: 
assumes "(∀h. (f (x + h) - f x) = D*h)"
shows "DERIV f x :> D"
proof cases
assume notzero:  "∀h. h ≠ 0"
have cs1: "(λh. (f (x + h) - f x) / h) -- 0 --> D" using assms notzero by auto
from this DERIV_def show ?thesis by auto

From the assumptions, I can easily prove the lemma by taking the limit then using DERIV_def. For this I have to assume that h ≠ 0. Continuing with the proof by cases I have to show that even when h = 0, the goal is true, however this can't be done when h = 0 as the assumption becomes 0 = 0. The lemma becomes trivial.

Is there a way I can prove the goal, which is this case is that f has derivative D at x, without the additional assumption that h ≠ 0?

edit: After further research, I came across the use of elimination rules in Isabelle which may be helpful. Also, I understand that the lemma is correct as the if the function is continuous, then the derivative at 0 also exists. I have been searching for the correct use and implementation of the the above information. How can I improve my search, and where should I be looking?

A K
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  • At least your assumption `∀h. h ≠ 0` looks wrong at this place, since you assume that every `h` is different from 0, which is obviously wrong as `h` itself may be 0. You might first want to fix some `h` and then perform the case analysis on `h ≠ 0`. – René Thiemann Feb 27 '15 at 08:43
  • Thank you for your comments @RenéThiemann. Could you just explain how I would perform the case analysis that `h ≠ 0`? – A K Feb 28 '15 at 00:37

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