Given a bijective function $F: \mathbb{F}_2^n \rightarrow \mathbb{F}_2^n$.
The entry of the Difference Distribution Table (DDT) at row $\alpha$ and column $\beta$ is defined as
$$DDT_{F}(\alpha,\beta) = \delta_F(\alpha, \beta) = |\{ x \in \mathbb{F}_2^n | F(x) + F(x+\alpha) = \beta \}|$$
I need to show that
$$DDT_{F}(\alpha,\beta) = DDT_{F^{-1}}(\beta, \alpha)$$
with
$$DDT_{F^{-1}}(\beta, \alpha) = \delta_{F^{-1}}(\beta, \alpha) = |\{ x \in \mathbb{F}_2^n | F^{-1}(x) + F^{-1}(x+\beta) = \alpha \}|$$
Unfortunately, I do not know how to start to show the equality, but I assume substitution is the way to go. Can you please give me a hint?
I tried the same approach I used to show that this symmetric property does also hold for the Linear Approximation Table (LAT), i.e.
$$LAT_{F}(\alpha,\beta) = LAT_{F^{-1}}(\beta, \alpha)$$
with
$$LAT_F(\alpha,\beta) = \widehat{F}(\alpha,\beta) = \sum_{x \in \mathbb{F}_2^n} (-1)^{<\alpha,x> + <\beta,F(x)>}$$
$$LAT_{F^{-1}}(\beta,\alpha) = \widehat{F^{-1}}(\beta,\alpha) = \sum_{x \in \mathbb{F}_2^n} (-1)^{<\beta,x> + <\alpha,F^{-1}(x)>}$$
For the LAT, the trick was to substitute. Since $F$ is a permutation and the sum goes over all elements in $\mathbb{F}_2^n$, substitution only changes the order in which the elements in $\mathbb{F}_2^n$ are processed.
$$x = F(y)$$
This leads to:
$$\widehat{F^{-1}}(\beta,\alpha) = \sum_{x \in \mathbb{F}_2^n} (-1)^{<\beta,x> + <\alpha,F^{-1}(x)>} = \sum_{y \in \mathbb{F}_2^n} (-1)^{<\beta,F(y)> + <\alpha,y>} = \widehat{F}(\alpha,\beta)$$
Unfortunately, substituting $x = F(y)$ did not solve my issue with the DDT.
