$P \ne NP$: a proof relating complexity theory to block ciphers I started thinking about P vs NP after reading another question on this stack exchange. Here I propose a proof that relates P vs NP to the existence of a secure block cipher in the elf model.
Let's say I have a (normal) Turing machine and I want to model a nondeterministic machine with it. According to certain models, a nondeterministic Turing machine can branch at each decision to multiple instances of itself. If I have an input $N$ of $n$ bits, let's say I must make a decision for each of $n$ bits: that's $2^n$ possible branches, operating in parallel. Since  a normal Turing machine must loop through every state of $N$, operating on each independently, wouldn't that mean $P \ne NP$ since I can construct a problem in $NP$ which requires me to operate on all possible $N$.
If we consider a an ideal block cipher we can say there exists a construction which requires you to operate on every possible input (consider the elf model). In the elf model, one must either form a lookup table of the entire mapping ($2^b$ blocks where $b$ is the block size) or iterate on every possible block. Therefore, if there exists a secure mode of operating on multiple blocks in the elf model, there must exist a problem in $NP$ which cannot be performed without iterating over every possible $N$. Since secure modes of operations are known to exist (if a secure block cipher exists), and the elf model proves a secure block cipher exists, this would prove $P \ne NP$.
If I can create a cipher which requires me to solve a cipher with a smaller block size (a perfect block cipher), I can force the decision to depend on every input bit, making the security rely only on obtaining perfect diffusion among the possible states, which many ciphers do already. This way I can create a secure block cipher from another secure block cipher, of smaller block size. Effectively, every SP-network is proof that $P \ne NP$.
Is this valid, or is there anything I'm missing? Is there anything unclear about my reasoning?
and the elf model proves a secure block cipher exists

It is worth mentioning that lower bounds in computationally limited settings do not &quot;lift&quot; [1] to lower bounds in computationally stronger settings. For example, there are lower bounds of $\Omega(\sqrt{|G|})$ on DL in the generic group model. This does not prove that DL is hard in &quot;real life&quot; though (and in fact, finite-field discrete log has $o(\sqrt{|G|})$ algorithms).
Secure block ciphers are not known to exist unconditionally. I don't know if your post precisely says this anywhere (this is the impression I get of the insight of your post though), but a proof that a secure block cipher exists immediately proves that $\mathsf{P}\neq \mathsf{NP}$.
This is because:

Secure Block ciphers are Pseudo-Random Permutations (PRPs)
PRPs imply One-Way Functions (is this HILL89? Idk its one of those old results)
The existence of One-Way Functions directly shows $\mathsf{P}\neq\mathsf{NP}$ (the problem of inverting a poly-time computable function is easily shown to be in $\mathsf{NP}$. The function being one-way means this can't be done in $\mathsf{P}$, and therefore we've found an $\mathsf{NP}$ language which is not in $\mathsf{P}$.


[1] There are areas of complexity theory which try to do exactly this though. Look up &quot;lifting theorems&quot; in communication complexity in particular, where lower bounds in weak models (generally things like decision trees) are &quot;lifted&quot; to lower bounds (for slightly different problems generally) in stronger models. This all being said, I do not expect lifting theorems to be useful for your application.