**First snake equation**.

We can show that the diagrams for the snake equation are equal in terms of feasibility relations by expanding the right hand side (RHS) and showing it equals the identity on \$$X\$$.

$\left(1_X \times \cup_X \right)\left(\cap_X \times 1_X\right) = 1_X$

Each of these definitions a function from the product of their inputs and outputs to \$$\mathbf{Bool}\$$.

The first part is cap and the identity: \$$(\cap_X \times 1_X ) : 1 \times X \nrightarrow X \times X^\mathrm{op} \times X\$$
$(\cap_X \times 1_X )((\bullet,x),(x',x^\ast,x'')) = (x^\ast \le x') \ \mathrm{and} \ (x \le x'')$
where \$$\bullet \in 1\$$.

Dually, we get \$$(1_X \times \cup_X ) : X \times X^\mathrm{op} \times X \nrightarrow X \times 1\$$
$(1_X \times \cup_X )((x',x^\ast,x''),(x,\bullet)) = (x'' \le x^\ast) \ \mathrm{and} \ (x' \le x)$

Composing, we get
$(1_X \times \cup_X ) (\cap_X \times 1_X )((\bullet,x),(x,\bullet)) = x \le x'' \le x^\ast \le x' \le x$
where the composite is true only if we can find one \$$x'',x^\ast,x'\$$ in \$$X\$$ such that \$$x\le x\$$, but this is always true since we can set \$$x = x' = x'' = x^\ast\$$.

Therefore \$$(1_X \times \cup_X ) (\cap_X \times 1_X ) \simeq 1_X\$$. Equality holds if \$$X \otimes 1 = X = 1 \otimes X\$$ because \$$(1_X \times \cup_X ) (\cap_X \times 1_X ) : 1\otimes X \nrightarrow X \otimes 1 \$$.

**Second snake equation**

Same as above, but with opposite preorders (i.e., turn the arrows around).

$(1_{X^\mathrm{op}} \times \cap_{X^\mathrm{op}})((x,\bullet),(x',x^\ast,x'')) = x'' \le x^\ast \ \mathrm{and} \ x' \le x$

and

$(\cup_{X^\mathrm{op}} \times 1_{X^\mathrm{op}})((x',x^\ast,x''),(\bullet,x)) = x^\ast \le x' \ \mathrm{and} \ x \le x''$

Composing gives
$(\cup_{X^\mathrm{op}} \times 1_{X^\mathrm{op}})(1_{X^\mathrm{op}} \times \cap_{X^\mathrm{op}})((x,\bullet),(\bullet,x)) = x \le x'' \le x^\ast \le x' \le x$
for some choice of \$$x'',x^\ast,x'\$$ in \$$X\$$, but again this is always possible since we can choose \$$x = x' = x'' = x^\ast\$$.

This composite behaves the same as the identity on \$$X^\mathrm{op}\$$, the dual object of \$$X\$$.

**Edits:** Changed the name of the element in 1 from 1 to \$$\bullet\$$. Also see Yoav's comment since I only prove that the two profunctors are equal along their diagonal.