diff --git a/switch_chain_polynomial_mixing_v2.txt b/switch_chain_polynomial_mixing_v2.txt
index 9339724..c2abd2c 100644
--- a/switch_chain_polynomial_mixing_v2.txt
+++ b/switch_chain_polynomial_mixing_v2.txt
@@ -733,7 +733,7 @@ Lemma 3.3 (three-row variance) does the whole job.
 
 Partition Omega_d by recording:
 
-  (C1) the graph induced on V \ T, i.e. all edges with neither endpoint
+  (C1) the graph induced on V  T, i.e. all edges with neither endpoint
        in T;
   (C2) for each outside vertex w not in T, the number of edges from w
        to T.  This number is an integer in {0, 1, 2, 3}.
@@ -758,16 +758,16 @@ T).  So e is a constant of the cell, in {0, 1, 2, 3}.
 --------------------------------------------------------------------------
 
 We build a binary matrix with rows indexed by T (three rows) and columns
-indexed by V \ T (n-3 columns) plus, if e is 1 or 2, ONE extra column
+indexed by V  T (n-3 columns) plus, if e is 1 or 2, ONE extra column
 (indexed, say, by the symbol *).  The entries are:
 
-  * for w in V \ T and i in T: M[i, w] = 1 if edge iw is present,
+  * for w in V  T and i in T: M[i, w] = 1 if edge iw is present,
     else 0.  This is exactly what "the number of edges from w to T"
     records (its column sum);
 
   * the extra column, when present, encodes the internal edges of T:
     - e = 0: no extra column.  Row sums are d_i (each i has all its
-      degree coming from V \ T alone).
+      degree coming from V  T alone).
     - e = 3: no extra column.  Row sums are d_i - 2 (each i has two
       of its degree inside T).
     - e = 1: the sole edge is, say, ij.  Extra column is s = (s_i, s_j,
@@ -813,7 +813,7 @@ include the extra column iff e = 1 or e = 2).  Two directions:
      extra column from the internal graph.
 
   <- (matrix to graph): the outside edges come from the entries in
-     V \ T columns; the internal edges come from the extra column (or
+     V  T columns; the internal edges come from the extra column (or
      are deterministic when e = 0, 3).
 
 Every cell graph arises exactly once, and every fixed-margin matrix
@@ -837,7 +837,7 @@ matrices).  Uniform.
 Now identify what h_{ {j,k} }, h_{ {i,k} }, h_{ {i,j} } do, inside this
 cell, in matrix terms.
 
-Fix i in T, let {j,k} = T \ {i}.  Consider h_{ {j,k} }, i.e. E_{ {j,k} }
+Fix i in T, let {j,k} = T  {i}.  Consider h_{ {j,k} }, i.e. E_{ {j,k} }
 projected on {j,k}-fibers.
 
 Claim: within the cell, E_{ {j,k} } f = E[ f | S_i ], where S_i is
@@ -846,7 +846,7 @@ Claim: within the cell, E_{ {j,k} } f = E[ f | S_i ], where S_i is
 Reason:
 
   * External entries of row i determine the singleton assignment of
-    each w in V \ T to i: w is adjacent to i in the {j,k}-fiber iff
+    each w in V  T to i: w is adjacent to i in the {j,k}-fiber iff
     the entry M[i, w] is 1.
 
   * Whether M[i, *] is 1 (the extra column) is determined by the e
@@ -855,7 +855,7 @@ Reason:
     cell data + degree.
 
   * So the {j,k}-fiber is determined by: the cell data (fixed), plus
-    the external neighbors of i (row i of the matrix in the V \ T
+    the external neighbors of i (row i of the matrix in the V  T
     columns).
 
   * Conversely, fixing the {j,k}-fiber fixes the external neighbors of
@@ -1065,7 +1065,7 @@ Let's not be too abstract; concrete is better.  Here is the setup.
 
 Fix disjoint vertex pairs a = {i,j} and b = {k,l}.  Let
 
-    O  =  V \ (a union b)          (the other n - 4 vertices).
+    O  =  V  (a union b)          (the other n - 4 vertices).
 
 Partition Omega_d by recording:
 
@@ -1811,18 +1811,18 @@ Proof (Havel).  Fix a target order of neighbors and prescribe edges one
 vertex at a time, from lowest labelled vertex upward.  Suppose the
 vertices in {1, ..., v-1} have the correct neighbor sets in the current
 graph G, and let R = {v, ..., n}.  Prescribe the neighbors of v: the
-d_v^R vertices of R \ {v} of highest residual degree, ties by label.
+d_v^R vertices of R  {v} of highest residual degree, ties by label.
 If the current neighbors of v differ, there is a missing desired
 neighbor j and an undesired actual neighbor i, with
 d_j^R >= d_i^R.  Removing a possible edge ij subtracts the same amount
 from both residual degrees, so
 
-    |N_R(j) \ {i}|  >=  |N_R(i) \ {j}|.
+    |N_R(j)  {i}|  >=  |N_R(i)  {j}|.
 
-Since v ∈ (N_R(i) \ {j}) \ (N_R(j) \ {i}) (v is not adjacent to i or
-j inside R \ {v}, but v IS adjacent to i, its undesired neighbor),
-the set N_R(j) \ {i} must contain vertices not in N_R(i) \ {j}.
-Pick k in (N_R(j) \ {i}) \ (N_R(i) \ {j}).  Then:
+Since v ∈ (N_R(i)  {j})  (N_R(j)  {i}) (v is not adjacent to i or
+j inside R  {v}, but v IS adjacent to i, its undesired neighbor),
+the set N_R(j)  {i} must contain vertices not in N_R(i)  {j}.
+Pick k in (N_R(j)  {i})  (N_R(i)  {j}).  Then:
 
   vi, jk are edges;   vj, ik are NOT edges;   k is distinct from v, i, j.
 
@@ -1830,7 +1830,7 @@ So {vi, jk} -> {vj, ik} is a valid SWITCH.  It fixes every other edge
 of v and increases the number of v's correctly placed neighbors by 1.
 
 Repeat until v has the prescribed neighbors.  Then freeze v (its degree
-and neighbors are now correct, and future switches involve only R \ {v}).
+and neighbors are now correct, and future switches involve only R  {v}).
 Proceed to the next vertex.  The procedure terminates: each frozen
 vertex is prescribed independently of the starting graph.