diff --git a/kadison_kaplansky_projection_counter_v2.txt b/kadison_kaplansky_projection_counter_v2.txt
index ff43de1..d7be52f 100644
--- a/kadison_kaplansky_projection_counter_v2.txt
+++ b/kadison_kaplansky_projection_counter_v2.txt
@@ -401,7 +401,7 @@ unitaries.  Two products of U with U* are used.
 
 because each t_j on the identity contributes the identity, and
 the N contributions sum to N.  So ||U||^2 = N and the operator
-inequality A = p U*U p <= N p holds immediately.
+ineuality A = p U*U p <= N p holds immediately.
 
    But A is NOT equal to N p.  U* U is an N-by-N matrix over
 C_r*(K) whose (i,j) entry is lambda_K(t_i^{-1} t_j).  Its
@@ -1127,7 +1127,7 @@ Constants are now set once and never altered again:
 
    C_*, C_bullet          from the geometry of the voltage cover;
    m                      =  an integer > L(201 C_*);
-   E_0, A_*, kappa        satisfying the displayed inequalities;
+   E_0, A_*, kappa        satisfying the displayed ineualities;
    theta_0                =  small radius for the expansion (7.6);
    ell                    =  the last free parameter, chosen large.
 
@@ -1215,7 +1215,7 @@ so that:
          Then ||T_P - Pi O Pi||  <=  eta  in the full regular
          norm.  Both operators have LAYER WIDTH ONE (only
          |i-k| <= 1 is nonzero), so their central Fourier
-         fields satisfy the same inequality almost everywhere and
+         fields satisfy the same ineuality almost everywhere and
          ||T_P(t)|| <= eta outside 0 < t < delta.
 
 The point of (G3) is that T_P is a FINITE sum of group
@@ -1301,7 +1301,7 @@ averaged quadratic form satisfies |<A_x xi, xi>| <= ||P_x A P_x||
 ||xi||^2.  Since the row and column coefficient sums of A are
 bounded by R and the same bounds hold after averaging, the scalar
 Schur test gives ||A - A_x|| <= C x^{-B} R.  Density plus the
-quadratic form identity give the claimed two-sided inequality.
+quadratic form identity give the claimed two-sided ineuality.
 Do nothing new for the amplification: the argument never touched
 the auxiliary index set, only the deck indices of F.
 
@@ -2456,7 +2456,7 @@ SPECTRUM OF C.  Since C = C* and C = 1 on F_D while C <= 1 -
 chi + alpha on (1 - F_D)H, the spectrum of C lies in
 [-||C||, 1 - chi + alpha] together with the eigenvalue 1 at F_D.
 
-SPECTRUM OF T.  Weyl's inequality: the spectral slope of T is at
+SPECTRUM OF T.  Weyl's ineuality: the spectral slope of T is at
 most ||T - C|| away from the spectrum of C.  So
 
    spectrum(T)  contained in
@@ -2597,7 +2597,7 @@ On the OTHER HAND, on p_0 H the spectral split gives
    < T xi, xi >  >=  (1 - 3 alpha) ||xi||^2.
 
 Since 1 - 3 alpha > 1 - chi + 3 alpha (with alpha = chi/1000), no
-nonzero xi can satisfy both inequalities.  Therefore
+nonzero xi can satisfy both ineualities.  Therefore
 p_0 H  ∩  K  =  {0}.
 
 KERNEL-COMPLEMENT SIZE.  Polar-decompose the restriction map