diff --git a/trace_reconstruction_anchor_induction_v2.txt b/trace_reconstruction_anchor_induction_v2.txt
index 1769544..146bd41 100644
--- a/trace_reconstruction_anchor_induction_v2.txt
+++ b/trace_reconstruction_anchor_induction_v2.txt
@@ -141,11 +141,12 @@ strings over an arbitrary alphabet, assuming the surviving symbols are
 observed exactly.
 
 Sanity check on sizes. Fix p = 1/2 and n = 10^9. Then log n is about
-20.7, (log n)^3 is about 8900, log log(2n) is about 3.05,
-(1 + log log(2n))^6 is about 1250, and the product is about 1.1 * 10^7.
-So the theorem promises fewer than exp(2.2 * 10^7) traces. That is a
-finite number, and the shape of the exponent is the point. A trivial
-bound is 2^n, which here is 10^{3 * 10^8}: vastly larger.
+20.7, (log n)^3 is about 8900, log log(2n) is about 3.06,
+(1 + log log(2n))^6 is about 4500, and the product is about 4.0 * 10^7.
+With p^{-1} = 2 included, the theorem promises fewer than
+exp(8 * 10^7) traces. That is a finite number, and the shape of the
+exponent is the point. A trivial bound is 2^n, which here is
+10^{3 * 10^8}: vastly larger.
 
 Vocabulary. A function of n is polynomial if it is at most n^C for some
 fixed C. It is quasipolynomial if it is at most exp((log n)^C) for some
@@ -837,7 +838,7 @@ such that
 
     | prod_i F(l_i) - prod_i G(l_i) |  >  Lambda.
 
-Sketch of proof. Introduce the deficits
+Proof. Introduce the deficits
 
     m_l = max( |F(l)|, |G(l)| )               (so 0 <= m_l <= 1)
     g_l = min( 50, E^{-1} log(1/m_l) )
@@ -860,16 +861,26 @@ the deficit budget
 
     G_cal + Q <= 1.003.
 
-The proof is a Riemann-sum approximation to the Poisson integral.
-Concretely, for each l,
+Proof of the deficit budget, in two parts.
 
-    g_l + q_l <= min( 100, E^{-1} log( 2 / |J(exp(-a + i a l h))| ) )
+Continuous bound. Since J_diamond is a polynomial, log|J_diamond| is
+subharmonic on the unit disk. Evaluating at the interior point z_0 (push
+the disk radius just past exp(-a) and let it return; capping at 100E can
+only lower an average) and using the pivot bound gives
 
-up to the truncation correction, and the Poisson integral of that capped
-quantity is at most 1.0025. The rest is bookkeeping. The Poisson density
-in the rescaled variable x = l h, for the disk of radius exp(-a), is at
-least (1 - 2a)/(pi (1 + (x - v_0^*)^2)), and the grid sum differs from
-the integral by o(1) uniformly, by layer cake and bounded variation.
+    Poisson average of log( 2 / |J_diamond| )  <=  log 2 + 1.0006 E,
+
+which is at most 1.003 E for E >= 300. Under theta = a x, the Poisson
+kernel at z_0 has density at least (1 - 2a)/(pi (1 + (x - v_0^*)^2)),
+so the continuous integral dominates the Riemann sum that produces
+G_cal + Q.
+
+Discretization. By layer cake, f(x) = int_0^{infinity} 1_{f(x) > c} dc
+for f(x) = min(100, E^{-1} log(2/|J_diamond(exp(-a + i a x))|)). A
+polynomial with H_* index terms meets a fixed positive modulus at most
+O(H_*) distinct values of x. So each level set f > c is a union of
+O(H_*) intervals. Replacing each interval's length by the grid sum of
+mesh h adds O(H_* h) = o(1) uniformly in n.
 
 Now the combinatorial core. For every integer j'shift with
 1 <= j'shift <= S/h, every fractional vertex cover of the graph on the
@@ -956,8 +967,8 @@ while
 
     T_lim(w-1) = 6 exp((w-1)^2/20) - .1
 
-equals about 5.90, 6.52, 7.41, 8.73, 10.51. Each exceeds the
-corresponding maximum.
+equals about 5.90, 6.21, 7.23, 9.31, 13.25. Each exceeds the
+corresponding maximum (4.96, 5.79, 6.94, 8.26, 9.71).
 
 Now the contradiction. Put R = G_cal/M and T_0 = T_lim(v_0^*), which is
 at least T_lim(w-1). Then 2R < T_0, and there exists an integer s with
@@ -1155,7 +1166,7 @@ Moreover, for all real omega and all g_s >= 0,
 
     |K(g, omega)| <= exp( - epsilon t (g_1 + ... + g_r) ).        (11.1)
 
-Sketch of proof. Scale:
+Proof. Scale:
 
     z        = (Y_0 - g_bar) sqrt(r) / sigma
     rho      = t sigma / sqrt(r)
@@ -1270,7 +1281,7 @@ H_0, with
     int_{|lambda|_inf > C_3 R} |J_cal_hat(lambda)| dlambda
        <= (1/4) exp(-20B).                                        (12.2b)
 
-Sketch of proof. Continue the integral formula for K to complex
+Proof. Continue the integral formula for K to complex
 g = x - i y, with x and y real vectors in R^{d'} and the last coordinate
 set to zero. Using Re Q(x - i y) = Q(x) - Q(y), one gets
 
@@ -1575,7 +1586,13 @@ both sides contribute 1), we get
                     * G(varsigma).
 
 The powers of z_0 and z_e reorganize the inner sum into exactly
-G(varsigma) times an overall factor.
+G(varsigma) times an overall factor, via the elementary identity
+
+    z_0^{i_1} prod_{e=1}^{k-1} (z_0 z_e)^{i_{e+1}-i_e-1}
+       = z_0^{i_k-k+1} prod_{e=1}^{k-1} z_e^{i_{e+1}-i_e-1},
+
+which shows where each power of z_0, each power of a z_e, and each
+factor of G(varsigma) comes from.
 
 Modulus of the prefactor. We have