Finite ∨ finite is finite in any von Neumann algebra, comparisons only

KR 6.3.8 (p, q finite ⇒ p ∨ q finite) can be proved with no trace, no center-valued dimension and no countable additivity. After the parallelogram law, take e ⊥ f finite and use the comparison theorem to get f ≾ e on a central piece. If e + f is infinite, cut down to a nonzero properly infinite central piece and halve twice, giving y₁, y₂, y₃ ~ x = e + f, orthogonal. Key estimate: e = e₁ + e₂ with eᵢ ≾ y ∧ e. Proof: three parallelogram laws, plus finite cancellation r ~ r', c ~ c' below them ⇒ r − c ~ r' − c'. Then 'three halves are too many': for orthogonal s₀, s₁, s₂ ≤ e finite, each with e a sum of two pieces ≾ sᵢ, compare s₁ and s₂ centrally. On the piece where s₂ ≾ s₁, e ≾ s₂ + s₁ ≤ e, so by finiteness s₀ vanishes there, and then e does too. Symmetrically on 1 − z, so e = 0.