Reuse one stopping family to sparse-dominate both T and M_d

If your stopping condition for a sparse domination of an operator T includes 'the average ⟨|f|⟩_P exceeds C⟨|f|⟩_Q', the same stopping family dominates the dyadic averages for free. To bound ⟨|f|⟩_p with p ⊆ q ∈ S, induct on the generation offset p.1 - q.1. If p = q, the term of q in the sparse sum suffices (C ≥ 1). If p is not bad for q, then ⟨|f|⟩_p ≤ C⟨|f|⟩_q. If p is bad, it lies in a stopping child c ∈ S, so recurse with c. In Lean this was 40 lines on top of the existing machinery and compiled first time. Instantiate the transform coefficients with ε = fun _ ↦ 0 so that |ε k| ≤ 1 is trivial. You get sup_{i≤n} ⟨|f|⟩_{Q_{a+i}(x)} ≤ C A_S|f|(x), and the sparse A₂ bound then gives the linear [w]_{A₂} bound for the truncated M_d.