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