In Proposition 7.2.4 of Loday and Vallette's Algebraic Operads, they prove that the Koszul dual operad of a quadratic operad $\mathcal{P}(E,R)$, with generators E finite dimensional in each arity and quadratic relations R, is given by
$$\mathcal{P}(s^{-1}\mathcal{S}^{-1}\otimes_H E^*, R^\perp)$$
where $s^{-1}$ is the desuspension operator, $\mathcal{S}^{-1}$ is the endomorphism operad of the one dimension vector space in degree -1, and $\otimes_H$ is the Haramard product. They do this by showing the Koszul dual operad is isomorphic (up to shift in degree) to a quotient
$$\mathcal{T}(F\otimes_H s^{-1}E^*)/(\Phi^{-1}( U\otimes_H \mathcal{T}(E^*) + \mathcal{T}(F)\otimes_H (s^2R)^\perp))$$
where $S^{-1}=\mathcal{T}(F)/(U)$ is a quadratic linear presentation, and $\Phi\colon\mathcal{T}(F\otimes_H s^{-1}E^*)\to \mathcal{T}(F)\otimes_H\mathcal{T}(s^{-1}E^*)$ is the obvious map.
I follow the proof up until this point, but then I don't understand the following statement:
Since U is quadratic and linear and since R is quadratic, this latter operad is quadratic. The space of relations, denoted $R^\perp$ is obtained from $(s^2R)^\perp$ by proper (de)suspension of the operations indexing the vertices of the trees and by induced sign rule
So are we just defining $R^\perp$ to be the space of relations? I had always thought this was genuinely an orthogonal subspace with respect to an appropriate inner product. Could someone expand on this choice of notation, or direct me to an extended proof of this proposition?