I proved P=NP: by showing that if it was not true then, for any boolean function $f$ which was neither $0,1$ or a constant function, it was possible to determine from the bits of $f$ whether $f$ is $0$ or $1$.
My proof:
let $f$ be any boolean function that is neither 0 nor 1. This means $f(x) \neq 0$ for every $x \in \{0,1\}$ and $f(x) \neq 1$ for every $x \in \{0,1\}$.
Now consider the 2-state Markov chain $M$ where the rows are labelled by the 4 binary strings that encode $f(0),f(1),f(00)$ and $f(11)$ and the columns are labelled by $0,1,00$ and $11$.
I have already shown that if $M$ is aperiodic then $P \neq NP$, but I would like to show that if $P=NP$ then $M$ is not aperiodic, by showing that there is a path with no cycle in $M$.
So I considered $p = f(0)f(0)f(0)f(1)$ and $q = f(0)f(0)f(0)f(0)$ and I will now prove that if $M$ is aperiodic then $p \in L(M)$ and $q \not \in L(M)$ (where $L(M)$ is the language of $M$)
Suppose that $M$ is aperiodic, so $M^k$ is not periodic for every $k$.
Now, we know that $M^k$ is not periodic for every $k$. But since $f$ is neither $0$ nor $1$ (by aperiodicity) then $f(0)^k = f(0) \neq f(0)f(0)$ and $f(0)^k = f(0) \neq f(0)f(0)f(1)$. So $f(0)$ is not contained in $f(0)f(0)f(0)f(1)$, and $f(0)f(0)f(0)f(0)$ is not contained in $f(0)f(0)f(0)f(1)$. So we have $p \in L(M)$ and $q \not \in L(M)$.
Is it right?