diff --git a/lattice_covering_n_log_n_v2.txt b/lattice_covering_n_log_n_v2.txt
index 21bc771..f99c4fb 100644
--- a/lattice_covering_n_log_n_v2.txt
+++ b/lattice_covering_n_log_n_v2.txt
@@ -265,12 +265,12 @@ CONSTANTS. The letters c and C always denote absolute positive
 constants, possibly changing from line to line. They never depend on
 the convex body K or on the growing dimension. The decimal exponents
 used later (0.56, 0.65, 0.69, 0.70, ...) are chosen only to satisfy
-strict inequalities; they are not optimal.
+strict ineualities; they are not optimal.
 
 ASYMPTOTICS. "Sufficiently large n" (or sufficiently large block size
 b) means beyond an absolute threshold that does not depend on K. Some
 lemmas fix an absolute integer B large enough for a finite list of
-inequalities to hold; that B is chosen once, before n grows, and stays
+ineualities to hold; that B is chosen once, before n grows, and stays
 absolute throughout.
 
 TWO SMALL INEQUALITIES WE USE REPEATEDLY.
@@ -380,7 +380,7 @@ v_0,...,v_D have norm 2DR < S. By (7.2) each slice K_{v_i} is nonempty,
 so pick x_i in K_{v_i}. There is an affine map T: R^D -> R^m with
 T(v_i) = x_i for all i. Replace K by its image under the determinant-one
 map (x,y) -> (x - T(y), y). This translates each horizontal slice by a
-fixed vector, so slice volumes and inequality (7.2) are unchanged. The
+fixed vector, so slice volumes and ineuality (7.2) are unchanged. The
 new body contains (0, v_i) for every vertex i; by convexity it contains
 (0,y) for every y in the simplex, in particular for every |y| <= 2R.
 
@@ -399,7 +399,7 @@ convexity gives
     (1 - m^{-2}) K_{y_a}  is contained in  K_y.             (7.4)
 
 Step 5: a body of prescribed volume inside the scaled slice. By
-Bernoulli's inequality (Section 6, item i), (1 - m^{-2})^m >= 1 - m^{-1}
+Bernoulli's ineuality (Section 6, item i), (1 - m^{-2})^m >= 1 - m^{-1}
 >= 1/2 for m >= 2. Hence by (7.2) the scaled slice (1 - m^{-2}) K_{y_a}
 has volume at least (1/4) vol_n(K) gamma_D(y_a). That scaled slice is
 convex and contains 0 (Step 2). Shrinking it about 0 by a further
@@ -485,7 +485,7 @@ the hypothesis max_a u_a <= eta_m,
         <= e^{-u_a} + C_R e^{-eta_m}
         <= (1 + C_R) e^{-u_a}.                          (7.9)
 
-Step 3: bound the failure probability of one a by Markov's inequality.
+Step 3: bound the failure probability of one a by Markov's ineuality.
 For a nonnegative random variable X, Prob(X > t) <= E X / t. Using
 (7.9) with t = e^{-u_a/2},
 
@@ -613,7 +613,7 @@ average to gbar^2, and the diagonal contributes E g^2. So
         <= 2^{-k} E_x g(x)^2
         <= 2^{-k} exp(2 b^{0.65}).                        (8.4)
 
-The last inequality is (8.1): g <= exp(b^{0.65}), so g^2 <=
+The last ineuality is (8.1): g <= exp(b^{0.65}), so g^2 <=
 exp(2 b^{0.65}) pointwise.
 
 Step 1.3: a controlled bad fraction. For fixed first-stage shifts, put
@@ -622,7 +622,7 @@ Step 1.3: a controlled bad fraction. For fixed first-stage shifts, put
     beta = 1 - |G|/|H|.
 
 Here beta is the FRACTION of base points on which A is bad. Chebyshev's
-inequality, together with gbar >= 1/2 and (8.4), gives
+ineuality, together with gbar >= 1/2 and (8.4), gives
 
     E_{w_1..w_k} beta <= 4 b^{1/5} 2^{-k} exp(2 b^{0.65})
                       <= exp(-2 c_* s).                    (8.5)
@@ -637,7 +637,7 @@ terms are at most (1/4) s log 2 once b is large; the remaining amount
 beats 2 c_* s. So for b beyond an absolute b_0, (8.5) holds, uniformly
 in s in the allowed range.
 
-Step 1.4: convert the average to a probability. Markov's inequality on
+Step 1.4: convert the average to a probability. Markov's ineuality on
 the nonnegative random variable beta gives
 
     Prob { beta > exp(-c_* s) } <= exp(-c_* s).            (8.6)
@@ -686,7 +686,7 @@ far, average over the fresh shift w = w_{k+j+1}:
 
 The second equality uses the bijection (x, w) -> (x, x - w) of H x H.
 
-Step 2.4: one bad step is rare. Conditional Markov's inequality applied
+Step 2.4: one bad step is rare. Conditional Markov's ineuality applied
 to (8.8) says
 
     Prob( log Phi_{j+1} > 2 log Phi_j + theta delta | previous shifts )
@@ -971,7 +971,7 @@ falling in this interval. Then N is binomial with
     E N >= c' h_b^{-1} b^{23/32} >= 2 b^{0.70}
 
 once b is large, because 23/32 > 0.70 leaves room. Its variance is at
-most its mean, so Chebyshev's inequality gives
+most its mean, so Chebyshev's ineuality gives
 
     Prob( N < b^{0.70} ) <= 4 / E N
         <= C h_b b^{-23/32} = O(b^{-1/20}).             (9.7)
@@ -997,8 +997,8 @@ every coordinate),
     | log F_b(t) - log F_b(t') | <= sum_j h_b^2 |t_j - t'_j|
         <= b h_b^2 / p_b = eta_b.
 
-This is the first inequality in (9.3). For the second inequality in
-(9.3), apply (9.0) at the anchor t_z and (9.3)-first inequality at the
+This is the first ineuality in (9.3). For the second ineuality in
+(9.3), apply (9.0) at the anchor t_z and (9.3)-first ineuality at the
 generic point t: the factor P_{b,t_z}(eps) is bounded by the folded
 ratio, giving the displayed sandwich. This proves (9.3).
 
@@ -1006,7 +1006,7 @@ Step B4: rounding preserves eligibility. Rounding a continuous point
 down to t_z changes log F_b by at most eta_b and each coordinate by
 less than 1/p_b. Since eta_b < (1/2) b^{0.56} and 1/p_b < (2 h_b^2)^{-1}
 for large b, the stricter continuous event from Step B1/B2 rounds INTO
-the eligible set E_b. Summing (9.3)-first inequality over all cells
+the eligible set E_b. Summing (9.3)-first ineuality over all cells
 gives
 
     e^{-eta_b} E_z F_b(t_z) <= integral_{[0,1]^b} F_b <= e^{eta_b} E_z F_b(t_z).
@@ -1034,13 +1034,13 @@ F_b(t_z)/p_b <= exp(5 b^{0.56}). Since 0.56 < 0.70, the product
         <= exp(5 b^{0.56} - c' b^{0.70}) <= exp(-c b^{0.70})
 
 for an absolute c > 0 and b sufficiently large (take c = c'/2). This is
-the first inequality in (9.4). The second inequality in (9.4) follows
-by (9.3)-second inequality applied at any t in Q_z. That completes the
+the first ineuality in (9.4). The second ineuality in (9.4) follows
+by (9.3)-second ineuality applied at any t in Q_z. That completes the
 proof of Lemma 9.1.
 
 Before going on, here is a short account of why these particular
 decimal exponents were chosen. They are not optimal; they only need to
-satisfy a chain of strict inequalities:
+satisfy a chain of strict ineualities:
 
     1/2 < 0.56,
     0.56 / 0.9 < 0.65 < 0.69 < 0.70 < 23/32.
@@ -1051,7 +1051,7 @@ Step B1). 0.56/0.9 < 0.65 makes the weight cap of the PREVIOUS block
 fit the sampler hypothesis (8.1) of the NEXT block, once we set
 b_next = b^{0.9} (see section 10). 0.65 < 0.69 < 0.70 leaves room
 between the weight cap, the number of "fair" bits, and the number of
-balanced coordinates; each is used in a different inequality. Finally
+balanced coordinates; each is used in a different ineuality. Finally
 0.70 < 23/32 is the numerical slack used in Step B2, and the strictness
 of 2(0.70) > 1 is what lets the atom bound (9.4) at the FIRST block
 outrun the logarithmic loss that appears in section 12 when we dilate
@@ -1151,7 +1151,7 @@ chosen into S with probability at least 2 alpha. Since the anchor z is
 eligible, there are at least b^{0.70} balanced coordinates (9.2b). Let
 K be the number of these balanced coordinates that are chosen into S.
 Then K is binomial with mean at least 2 alpha b^{0.70} and variance at
-most its mean. Chebyshev's inequality, using that b^{0.69} is at most
+most its mean. Chebyshev's ineuality, using that b^{0.69} is at most
 HALF the mean for large b, gives
 
     Prob( K < b^{0.69} ) <= 4 / (2 alpha b^{0.70})
@@ -1183,13 +1183,13 @@ for each eligible z. Then averaging the weighted mass over z in E_b:
         <= C b^{-1/2} E_z F_b(t_z) ind_{E_b}(z)
         <= C' b^{-1/2}.
 
-The last inequality uses (9.2), which says the weighted grid mass on
+The last ineuality uses (9.2), which says the weighted grid mass on
 E_b is 1 + O(b^{-1/20}) <= 2. So there is at least one W with the inner
 expectation <= C' b^{-1/2}; fix that W. Then define
 
     U_b = { z in E_b : q_W(z) <= b^{-1/20} }.
 
-Weighted Markov's inequality gives the explicit discarded mass
+Weighted Markov's ineuality gives the explicit discarded mass
 
     E_z F_b(t_z) ind_{E_b ∖ U_b}(z)
         <= C' b^{-1/2 + 1/20} = C' b^{-9/20}.               (9.12)
@@ -1470,7 +1470,7 @@ b = b_i, we have b_{i-1} <= b^{1/0.9} and
     log|G_{i-1}| = b_{i-1} log p_{i-1}
         = O( b_{i-1}^2 log b_{i-1} ) <= b^4.
 
-The first inequality uses 0.56/0.9 < 0.65, and the second has the slack
+The first ineuality uses 0.56/0.9 < 0.65, and the second has the slack
 2/0.9 < 4. The means fall into [1/2, 2] after another increase of B.
 We obtain usable sets U_i subset G_i, lines C_i subset G_i, and, for
 i < r, matrices W_i of size b_i by b_{i+1} with entries in F_{p_i}. The
@@ -1961,7 +1961,7 @@ Multiplying,
 Compare this to the target bound L_*^{k_1 + k_2 - 1} prod_{l} eta_{a(l)}.
 The average is smaller by a factor L_*.
 
-Step 6: control the failure probability. Markov's inequality: since the
+Step 6: control the failure probability. Markov's ineuality: since the
 integrand is nonnegative and has the average above, the probability
 that the integrand exceeds L_* times its average is at most 1/L_*. In
 other words, the probability that Z_q VIOLATES (11.1) for this particular
@@ -2006,7 +2006,7 @@ satisfies
 
 for a positive integer k, then (1 + 1/k) K + Gamma = R^n.
 
-Notice the two inequalities that make (11.2) work. On the right, k^{-n}
+Notice the two ineualities that make (11.2) work. On the right, k^{-n}
 is much smaller than 1. On the left, delta must be very small to begin
 with. The precise form ensures that the SMALLER the hole, the LESS
 dilation is needed. As such, a virtually-covered body needs almost no
@@ -2032,7 +2032,7 @@ Rearranging and using delta = 1 - |pi_Gamma(K)|:
 
     |pi_Gamma(K/k)| >= k^{-n} (1 - delta) > delta = 1 - |pi_Gamma(K)|,
 
-where the strict inequality is the hypothesis (11.2).
+where the strict ineuality is the hypothesis (11.2).
 
 Step 3: conclude the covering. For every torus point [x], the two sets
 pi_Gamma(K) and [x] - pi_Gamma(K/k) have measures summing to more than
@@ -2126,7 +2126,7 @@ Gaussian weight multiplied by rho:
 Here cmp is a benign absolute comparability constant from (12.1)
 combining c_sec and C_sec, and the m-dimensional volumes are matched to
 D_v gamma_D via the definition of D_h in (12.2a). The two load
-inequalities follow:
+ineualities follow:
 
     sum_{ell in P_y} u_{a(ell)} >= c_3 rho,                 (12.2c)
     max_{ell in P_y} u_{a(ell)} <= U_n := C rho exp(-c b_1^{0.70}),
@@ -2138,7 +2138,7 @@ Equation (12.2c) is the SUM OF LOADS along a pattern (must be large);
 (12.2d) is the MAXIMUM LOAD of any individual term (must be small).
 
 Step 5: three scale comparisons. We need (12.2c)-(12.2d) to fit the
-criteria of Lemma 7.2. There are three inequalities, all of the form
+criteria of Lemma 7.2. There are three ineualities, all of the form
 "small is small". First the load cap:
 
     U_n / m
@@ -2339,7 +2339,7 @@ lattice with tiny holes (Lemma 7.2); a simultaneous shear (Lemma 11.1)
 makes hole sets overlap as if independent; and finally completion
 (Lemma 11.2) plus dilation gives density O(n log n). The decimal
 exponents (0.56, 0.65, 0.69, 0.70, 23/32) are chosen only for strict
-inequalities, not optimality; the constant C is finite though large.
+ineualities, not optimality; the constant C is finite though large.
 
 ---------------------------------------------------------------------
 13. REFERENCES