Information equilibrium is an equivalence relation

Something for the math nerds. I've said it a couple times, but haven't actually shown the proof. However, it is true that information equilibrium is an equivalence relation. If we define the statement $A$ to be in information equilibrium with $B$ (which we'll denote $A \cong B$) by the relationship (i.e. ideal information transfer from $A$ to $B$):

$$\text{(1) }\;\; \frac{dA}{dB} = k \frac{A}{B}$$

for some value of $k$, then, first we can show that $A \cong A$ because

$$\frac{dA}{dA} = k \frac{A}{A}$$

$$1 = k \cdot 1$$

and we can take $k = 1$. Second we can show that $A \cong B$ implies $B \cong A$ by re-deriving the relationship (1), except moving the variables to the opposite side:

$$\frac{dB}{dA} = \frac{1}{k}\;\; \frac{B}{A} = k' \; \frac{B}{A}$$

for some $k'$ (i.e. $k' = 1/k$). Lastly we can show that $A \cong B$ and $B \cong C$ implies $A \cong C$ via the chain rule:

$$\text{(2a) }\;\; \frac{dA}{dB} = a \frac{A}{B}$$

$$\text{(2b) }\;\; \frac{dB}{dC} = b \frac{B}{C}$$

such that

$$\frac{dA}{dC} = \frac{dA}{dB}\; \frac{dB}{dC} = a b \; \frac{A}{B} \frac{B}{C}$$

$$\frac{dA}{dC} = k \; \frac{A}{C}$$

with information transfer index $k = a b$. That gives us the three properties of an equivalence relation: reflexivity, symmetry and transitivity.