Further, by enumerating all six possible permutations, it can be seen that only the φ with φ(f ) = f for f ∈B induces a score of > −∞.
we can rewrite to Here • (AB) is independent of φ since ∀f ∈Ã, f ∈ {n + 1, n + 3} : ∀f ∈Ã : • (BA) is independent of φ since ∀f ∈Ã, f ∈B : • (BB) is independent of φ.