$Fleft(mathbf r, frac{1}{i}nablaright)psi$ and $F(inabla, mathbf k)phi$ are FT's of each other if $psi$ and $phi$ are FT's of each other?

I’m reading Merzbacher’s Quantum Mechanics and he states the above (for analytic $F$).
To prove this, I could come up with the following, but am not sure if the last step is quite correct.

We have
begin{align}
F(inabla_{mathbf k’}, mathbf k)Big|_{mathbf k}left{ phi(mathbf k’) right}
&= frac{1}{left(2piright)^{3/2}} int psi(mathbf r); F(inabla_{mathbf k’}, mathbf k)Big|_{mathbf k} left{ e^{-imathbf{k’cdot r}} right}~mathrm d^3r
&= frac{1}{left(2piright)^{3/2}}int psi(mathbf r); F(mathbf r, mathbf k) e^{-imathbf{kcdot r}}~mathrm d^3r,
end{align}
where in the first step, I used that $phi$ is the FT of $psi$.

Now,
begin{alignat}{4}
frac{1}{left(2piright)^{3/2}} int F(inabla_{mathbf k’}, mathbf k)Big|_{mathbf k}left{
phi(mathbf k’) right} e^{imathbf{kcdot r}}~ mathrm d^3k
&= frac{1}{left(2piright)^3} &&int psi(mathbf r’); F(mathbf r’, mathbf k); e^{imathbf{kcdot (r-r’)}}~ mathrm d^3k; mathrm d^3r’
&= frac{1}{left(2piright)^3} &&intleft[ Fleft(mathbf r’, frac{1}{i}nabla_{mathbf r”}right)Bigg|_{mathbf r} left{ psi(mathbf r’) e^{imathbf{kcdot (r”-r’)}} right} right]~ mathrm d^3 k; mathrm d^3 r’
&= &&int Fleft(mathbf r’, frac{1}{i}nabla_{mathbf r”}right)Bigg|_{mathbf r} left{ psi(mathbf r’); delta^3(mathbf{r”-r’}) right}~ mathrm d^3r’,
end{alignat}

where in the second equality I used that $F(mathbf r’, mathbf k); e^{imathbf{kcdot r}} = F(mathbf r’, frac{1}{i}nabla_{mathbf r”})Big|_{mathbf r} left{ e^{imathbf{kcdot r”}} right}$.

Question: Can I now just say that the RHS equals $$Fleft(mathbf r”, frac{1}{i}nabla_{mathbf r”}right)Bigg|_{mathbf r} left{ psi(mathbf r”) right} = left[Fleft(mathbf r, frac{1}{i} nablaright)psiright](mathbf r),$$ completing my proof?

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Edit: As pointed by Valter in his comments, $F$ should be such that there are no commutativity issues, for instance between $nabla_{mathbf k’}$ and $mathbf k’$.