I know that if we have a cipher that makes only linear transformations (let's say a bunch of $XOR$s), we break it simply by writing a system of equations with $\oplus$ operation starting from the cipher's scheme.
In the end, we'll have a system of $n$ equation involving the $n$ plaintext bits, the $n$ ciphertext bits, and the $n$ key bits holding with probability: $\Pr(1)$ and we can solve for the $n$ key bits.
Why with the presence of S-boxes is this not possible in the same way?
Thinking about $DES$, couldn't we just reverse the S-Boxes (which at the end are only lookup tables) to write the same system of equations? In that case, the relations won't be again deterministic? (holding with $\Pr(1)$)
What is the precise meaning of non-deterministic transformation and how is it applied here?
Could you please make me a counter-example of what I'm saying, just to see why this approach couldn't be used as it is, with the presence of S-Boxes?