diff --git a/metric_k_median_threshold_v2.txt b/metric_k_median_threshold_v2.txt
index 7bb5629..b6b41ac 100644
--- a/metric_k_median_threshold_v2.txt
+++ b/metric_k_median_threshold_v2.txt
@@ -1,5 +1,5 @@
 THE APPROXIMATION THRESHOLD FOR METRIC k-MEDIAN
-A complete proof, written for a motivated high school reader
+A complete proof
 ================================================================
 
 CONTENTS
@@ -24,9 +24,7 @@ PREFACE
 -------
 
 This document proves one theorem and its matching hardness bound.
-It is written to be read straight through by a high school student
-who is comfortable with fractions, averages, and the idea of a
-limit. The proof is complete: every lemma is either proved from
+The proof is complete: every lemma is either proved from
 scratch or is a well-known external result stated precisely.
 
 The one sentence version:
@@ -470,7 +468,7 @@ For each representative s, define
   P_s = F_s                                    if L_s = infinity;
   p_s = y(P_s);
   c_s = 1 - p_s;      this is the "deficit" of s;
-  A_s = F_s \ P_s;    this is the "far part" of F_s.
+  A_s = F_s ∖ P_s;    this is the "far part" of F_s.
   U   = pieces outside every P_s.
 
 DISJOINTNESS. If P_s and P_t share a piece i (s != t), then
@@ -764,8 +762,8 @@ So D/R_s <= M V_j / (delta T_1 V_j) = alpha <= 1/4 < 1.
 
 Compare F_s (used to serve s) with F_j (used to serve j):
 Both F_s and F_j are "nearest-unit" sets, both of unit total mass.
-The masses of F_s \ F_j and F_j \ F_s are equal. Each piece in
-F_s \ F_j is at distance <= R_s + D from j. Each piece in F_j \ F_s
+The masses of F_s ∖ F_j and F_j ∖ F_s are equal. Each piece in
+F_s ∖ F_j is at distance <= R_s + D from j. Each piece in F_j ∖ F_s
 is at distance >= R_s - D from j (because the nearest-unit
 assignment at s uses all closer mass first). Comparing these residual
 costs shows
@@ -1155,7 +1153,7 @@ bound of 0 charge, which is more than enough.
 
 Oh! I think the point is a subtlety. Perhaps the paper means: regardless of whether charge is actually H_0 or 0, we claim I <= -H_0 and incoming charge <= H_0 - 0 = H_0 (if somehow the positive part were H_0 rather than 0). But since positive part is 0, the bound I <= -H_0 is even stronger than needed.
 
-Actually reading again: "Equation ~\eqref{eq:scalar-type-zero} gives
+Actually reading again: "Equation ~(eq:scalar-type-zero) gives
 $I_{s_i}(0;q,Z)\le-H_0$, which covers its incoming charge."
 
 Hmm. So the claim is I <= -H_0 and the incoming charge is <= H_0