From this link:
Where equation 1 is:
I can probably brute-force this by explicitly calculating this quantum circuit's effective 4x4 matrix and seeing that its equivalent to this teleportation operation - but what does this have to do with X=HZH? Where is x used here? Is X somehow equivalent to a cNOTgate? X, Z and H are all one-qubit gates...so I don't understand what they mean.
Here are three equivalent circuits:
The second equation can be understood from this related answer about the fact that $CZ_{1 \rightarrow 2} = CZ_{2 \rightarrow 1}$, where first index is the control qubit and the second index is the target qubit. The first equation is similar to this related answer, but let's prove it explicitly where we will use $HZH = X$:
\begin{equation*} I \otimes H \; (CZ_{1 \rightarrow 2}) \; I \otimes H= \\ = I \otimes H \; (|0\rangle \langle 0 | \otimes I + |1\rangle \langle 1 | \otimes Z) \; I \otimes H =\\ = |0\rangle \langle 0 | \otimes H H + |1\rangle \langle 1 | \otimes HZH = \\ = |0\rangle \langle 0 | \otimes I + |1\rangle \langle 1 | \otimes X = CNOT_{1 \rightarrow 2} \end{equation*}
because $HH = I$ and $HZH = X$.
