If we take \$$\mathcal{V} \$$ to be \$$\textbf{Bool} := (\mathbb{B}, \le_{Bool}, \text{true}, \wedge) \$$
and \$$\mathcal{X}(m, n) := m \le_X n \$$, \$$\mathcal{Y}(m, n) := m \le_Y n \$$, \$$\Phi(m, n) := m \le_{\Phi} n \$$ , then

$\mathcal{X}(x' , x) \otimes \Phi(x, y) \otimes \mathcal{Y}(y, y' ) \le_{Bool} \Phi(x' , y' )$

becomes

$( x' \le_X x ) \wedge \Phi(x, y) \wedge ( y \le_Y y' ) \le_{Bool} \Phi(x' , y' )$

Which is (almost) the feasibility relation mentioned in the introduction.

$\text{ if } x' \le_X x \text{ and } y \le_Y y' \text{ then } \Phi(x, y) \le_{Bool} \Phi(x' , y' )$

Does the difference in \$$\le \$$ and "then" matter? Is one of them wrong?

I suppose it is the fact that \$$\textbf{Bool} \$$ is quantale that makes them equivalent.

Help!