I was reading "Origins of the Calculus of Binary Relations" by Vaughan Pratt where he says "it consists of two components, a logical or static component and a relative or dynamic component" but it seems as if it should be possible to define the "static component" purely in terms of the "dynamic component". Is this the case?

As I understand it, and in more modernday terms, the question asks whether it is possible to define the operations $0, 1, \cap, \cup, \neg$ on $P(X^2)$ (operations belonging to the "static component") in terms of "dynamic" operations $\delta', \delta, \circ, \circ', ()^{op})$ where $\circ$ is relational composition (as an operator on $P(X^2)$): $$(R \circ S)(x, z) := \exists_y R(x, y) \wedge S(y, z);$$ $\circ'$ is De Morgan dual to $\circ$: $$(R \circ' S)(x, z) := \forall_y R(x, y) \vee S(y, z);$$ $\delta \in P(X^2)$ is the diagonal subset (the identity for $\circ$), $\delta'$ is its complement (the identity for $\circ'$), and $()^{op}$ is the relational converse: $$R^{op}(x, y) = R(y, x).$$ The answer is no. For example, take $X$ to be $\mathbb{R}$, and consider the relation $n$ where $n(x, y) \Leftrightarrow (x = y)$. Let $n'$ be the complement of $n$. I claim that the set $A = \{0, 1, \delta, \delta', n, n'\}$ is closed under the dynamic operations. It's clear that each of these elements is a fixed point under relational converse. It's also easy to check that $n \circ n = \delta$, and that $$n' \circ n = \delta' = n \circ n',$$ $$\delta' \circ n = n' = n \circ \delta',$$ $$n' \circ n' = 1 = n' \circ \delta' = \delta' \circ n' = \delta' \circ \delta',$$ $$1 \circ x = 1 = x \circ 1 \qquad x \in A, x \neq 0$$ $$0 \circ x = 0 = x \circ 0 \qquad x \in A$$ Since $A$ is closed under complementation and under $\circ$, it is also closed under $\circ'$. Thus $A$ is closed under all the dynamic operations. However, it is clearly not closed under intersection, since $n \cap \delta$ is the singleton $\{(0, 0)\}$. 




