If they were not, one would have by the second point above β = ±12α, β = ± α or β = ±2α. If the minus sign holds, then (α|β) is automatically < 0 and there is nothing to be proven. So we only need to consider the cases β = +12α, β = + α or β = +2 α. In the first case, α− β = β ∈ Δ, in the second case α − β = 0, and in the last case α − β = − α ∈ Δ so these three cases are in fact excluded by the assumption. We can therefore assume α and β to be linearly independent.