∗ ≃φ ⋆
by:
(X × X → X(x1, x2) ↦ ax1,x2) ≃φ (Y × Y → Y(y1, y2) ↦ by1,y2) :⇔ ∀ x1, x2 ∈ X : φ(ax1,x2) = bφ(x1),φ(x2) (ax1,x2 ∈ X for each x1 ∈ X and x2 ∈ X, by1,y2 ∈ Y for each y1 ∈ Y and y2 ∈ Y) ⇔ ∀ x1, x2 ∈ X : φ(x1 ∗ x2) = φ(x1) ⋆ φ(x2) ⇔ φ ∘ ∗ = ⋆ ∘ (φ × φ) Equivalence. No proof.