English
If nhds x and the two sets s,t have equal nhds at x, HasDerivWithinAt f f' s x is equivalent to HasDerivWithinAt f f' t x.
Русский
Если окружности x совпадают по соседству для s и t, то производные по s и t равны.
LaTeX
$$$\\text{hasDerivWithinAt_congr_set }\\; h:\\; (nhds\\,x).EventuallyEq\\; s\\; t \\Rightarrow \\mathrm{HasDerivWithinAt}(f,f',s,x) \\iff \\mathrm{HasDerivWithinAt}(f,f',t,x).$$$
Lean4
theorem hasDerivWithinAt_congr_set {s t : Set 𝕜} (h : s =ᶠ[𝓝 x] t) :
HasDerivWithinAt f f' s x ↔ HasDerivWithinAt f f' t x :=
hasFDerivWithinAt_congr_set h