I think that we have the following definitions for the profunctors involved in the first snake equation:

\begin{align} \lambda_{\mathcal{X}}^{-1}(a, \langle 1, b \rangle) &= \mathcal{X}(a, b) \\\\ (\cap\_\mathcal{X} \otimes 1\_\mathcal{X})(\langle 1, b \rangle, \langle \langle c, c' \rangle, c'' \rangle) &= \mathcal{X}(c', c) \otimes \mathcal{X}(b, c'') \\\\ \alpha_{\mathcal{X}, \mathcal{X}^\mathrm{op}, \mathcal{X}}(\langle \langle c, c' \rangle, c'' \rangle, \langle d, \langle d', d'' \rangle \rangle) &= \mathcal{X}(c, d) \otimes \mathcal{X}(d', c') \otimes \mathcal{X}(c'', d'') \\\\ (1\_\mathcal{X} \otimes \cup\_\mathcal{X})(\langle d, \langle d', d'' \rangle \rangle, \langle e, 1 \rangle) &= \mathcal{X}(d, e) \otimes \mathcal{X}(d'', d') \\\\ \rho_\mathcal{X}(\langle e, 1 \rangle, f) &= \mathcal{X}(e, f). \\\\ \end{align}

Using the definition of composition and the fact that \$$\mathcal{V}\$$ is a commutative quantale, we get the composition of the above profunctors to be

$\bigvee_{b, c, c', c'', d, d', d'', e} \mathcal{X}(a, b) \otimes \mathcal{X}(c', c) \otimes \mathcal{X}(b, c'') \otimes \mathcal{X}(c, d) \otimes \mathcal{X}(d', c') \otimes \mathcal{X}(c'', d'') \otimes \mathcal{X}(d, e) \otimes \mathcal{X}(d'', d') \otimes \mathcal{X}(e, f).$

This equation is equal to the idenity profunctor, \$$\mathcal{X}(a, f)\$$, if

$\bigvee_{y} \mathcal{X}(x, y) \otimes \mathcal{X}(y, z) = \mathcal{X}(x, z).$

But I don't quite know how to show this – have we already proven this equation?

*Edit:*
I think I know how to prove the last equation. We can use the two properties of the join:

- The join \$$j\$$ is an upper bound; so for any \$$y\$$ we have that \$$\mathcal{X}(x, y) \otimes \mathcal{X}(y, z) \le j\$$. If we pick \$$y\$$ to be \$$x\$$ then we get that \$$\mathcal{X}(x,z) \le j\$$.
- The join \$$j\$$ is the _least_ upper bound; so for any other upper bound \$$u\$$ we have that \$$j \le u\$$. Since \$$\mathcal{X}(x,y) \otimes \mathcal{X}(y,z) \le \mathcal{X}(x, z)\$$ for all \$$y\$$, then \$$\mathcal{X}(x,z)\$$ is an upper bound and \$$j \le \mathcal{X}(x, z)\$$.

Hence, the join is exactly \$$\mathcal{X}(x,z)\$$.