i’m by no mean a mathematician but i tried to make the statement clear.
In common words the statement say that if a component of a representation of the data is useless to compute a function, then it is useless in all representation where he is in.
I dont know if its true and my brain melt just reading what i ve wrote.
If the size of the start and arrival domain is not the same i ve already found counter exemple (even with just one less dimention for arrival domain). Of course if the function is bijective it will be impossible to met the hypothesis of existence of useless information.
Formalisation, every dot is one hypothesis:

$f: left{0;1right}^n rightarrow left{0;1right}^n $

$(g_j)_{jin(1;n)}$ n functions, all from $left{0;1right}^n$ to $left{0;1right}$

such as $B: (x_i)_{iin(1;n)} longmapsto (g_1((x_i)),….,g_n((x_i)))$ bijection

and $(forall jgeq 2: g_j((x_i)) = g_j((x_i’))) Rightarrow f((x_i)) = f((x_i’))$ (ie first componement is useless to compute f)

then let $(g_j)_{jin(1;n)}$ n functions, all from $left{0;1right}^n$ to $left{0;1right} $ with $g’_1$= $g_1$

and such as $B’: (x_i)_{iin(1;n)} longmapsto (g’_1((x_i)),….,g’_n((x_i)))$ bijection
then the result: $(forall j geq 2: g’_j((x_i)) = g’_j((x_i’)))Rightarrow f((x_i)) = f((x_i’))$ ($g_1$ is then also useless in another representation)
I will probably sleep soon, but i ll try to read answer tomorrow. Thank you for reading.