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?