diff --git a/boundary_graph_penrose_v2.txt b/boundary_graph_penrose_v2.txt
index fe7848a..f212298 100644
--- a/boundary_graph_penrose_v2.txt
+++ b/boundary_graph_penrose_v2.txt
@@ -27,7 +27,7 @@ Here E is the total energy, A is the area of the horizon, and the
 formula is written in units where Newton's constant and the speed of
 light are both 1.
 
-The Penrose inequality generalises this to spacetimes that are not
+The Penrose ineuality generalises this to spacetimes that are not
 stationary. It says that the area of a horizon always bounds the
 total mass from below: whatever happens in a collapse, the total
 mass in the spacetime is at least the mass that would be needed to
@@ -36,7 +36,7 @@ gravity can never be "undone": once you have made a horizon of
 area A, the total mass of the resulting spacetime can never be less
 than sqrt(A/(16 pi)).
 
-Penrose conjectured the inequality in 1973. Proving it is one of the
+Penrose conjectured the ineuality in 1973. Proving it is one of the
 great open problems of mathematical general relativity. It has been
 proved in two very important specialised cases: for time-symmetric
 or "Riemannian" initial data, and for the four-dimensional
@@ -49,7 +49,7 @@ carefully chosen deformation.
 
 1.2 The exact statement
 
-To state the Penrose inequality we need four ingredients:
+To state the Penrose ineuality we need four ingredients:
 
    (I)  An initial data set: a triple (M, g, K), where M is a
         three- or four-dimensional manifold (the "spatial slice"),
@@ -70,7 +70,7 @@ To state the Penrose inequality we need four ingredients:
         infinity, and that describe the total energy and momentum
         of the slice.
 
-With these in hand, the spacetime Penrose inequality states that,
+With these in hand, the spacetime Penrose ineuality states that,
 for suitable data obeying an energy condition, the total energy E
 satisfies
 
@@ -96,14 +96,14 @@ dimension.
 
 1.3 Why it is hard
 
-The difficulty of the Penrose inequality comes from three facts
+The difficulty of the Penrose ineuality comes from three facts
 stacked on top of one another.
 
 First: the horizon is the boundary of a trapped region, and trapped
-regions are defined by a nonlinear inequality on a certain first-
+regions are defined by a nonlinear ineuality on a certain first-
 order quantity called the expansion. There is no simple equation
-for the horizon; it obeys an inequality, not an equality, and it can
-sit anywhere inside the region where the inequality is tight.
+for the horizon; it obeys an ineuality, not an equality, and it can
+sit anywhere inside the region where the ineuality is tight.
 
 Second: the ADM mass E is defined only at spatial infinity, while
 the horizon sits somewhere in the interior. There is no obvious
@@ -129,7 +129,7 @@ spherical case), and under a "suboptimal-constant" relaxation
 
 1.4 What this paper does, in five lines
 
-The authors prove a boundary version of the Penrose inequality in
+The authors prove a boundary version of the Penrose ineuality in
 three and four spatial dimensions. They start with an initial-data
 set exterior whose inner boundary is strictly trapped. They replace
 it, via a highly controlled conformal-plus-graph deformation, by a
@@ -138,8 +138,8 @@ strictly negative mean curvature at its inner boundary, (c) has area
 bounded below by the original area on every enclosing cut, and (d)
 has ADM energy no larger than the original, up to an error that
 tends to zero. Once these four properties hold, the Riemannian
-Penrose inequality (Huisken-Ilmanen, Bray, Bray-Lee) applies to the
-new metric, and the mass-area inequality passes back to the
+Penrose ineuality (Huisken-Ilmanen, Bray, Bray-Lee) applies to the
+new metric, and the mass-area ineuality passes back to the
 original data by letting the error tend to zero.
 
 The word "boundary" is important: the new metric keeps a genuine
@@ -241,14 +241,14 @@ The import of the constraint equations is this: the physical
 quantities mu and J are integrals of motion, so they cannot be
 chosen freely. Any pair (g, K) we invent must satisfy the
 constraints, which couples the metric and the tensor in a highly
-nonlinear way. This is one of the reasons the Penrose inequality is
+nonlinear way. This is one of the reasons the Penrose ineuality is
 hard: a naive "just change the metric to make it nicer" approach
 runs into the constraint system.
 
 
 2.3 Energy conditions
 
-For the Penrose inequality to have any hope of being true, the data
+For the Penrose ineuality to have any hope of being true, the data
 must obey some energy condition. Otherwise you can have negative
 mass, which would immediately violate any mass-lower-bound.
 
@@ -273,7 +273,7 @@ Riemannian Penrose theorem.
 
 2.4 Asymptotic flatness and ADM mass
 
-The Penrose inequality lives in an asymptotically flat exterior: a
+The Penrose ineuality lives in an asymptotically flat exterior: a
 spacetime that looks like flat Minkowski space very far from the
 horizon. There is one distinguished end, a "spatial infinity" where
 the geometry settles down. In three and four dimensions we say
@@ -321,7 +321,7 @@ case with K = 0 the momentum vanishes automatically.
 
 The ADM mass is a key piece of the puzzle: it is the "total energy
 at infinity" of the slice. It only depends on the behaviour at
-spatial infinity, not on the interior. The Penrose inequality is a
+spatial infinity, not on the interior. The Penrose ineuality is a
 bound relating this boundary quantity to another boundary quantity,
 the horizon area A, with the interior providing the obstruction.
 
@@ -400,7 +400,7 @@ the new metric would risk equalling zero.
 2.6 Enclosing cuts and enclosing area
 
 A crucial definition: what counts as "the horizon" for the Penrose
-inequality, and how do we measure its area? The paper allows a very
+ineuality, and how do we measure its area? The paper allows a very
 general notion so that the answer is robust.
 
 An exterior Omega is a connected orientable n-manifold with a
@@ -469,7 +469,7 @@ The key points:
    theta_+ = 0. A future trapped surface has theta_+ < 0.
 
    Scalar curvature. R_g is the scalar curvature of the metric g.
-   In the Riemannian Penrose inequality we need R_g >= 0.
+   In the Riemannian Penrose ineuality we need R_g >= 0.
 
    Einstein constant. Mathematically one usually sets 8 pi = 1.
    The 3D part of the paper sometimes writes mu, J with explicit
@@ -486,7 +486,7 @@ PART 3. THE STRATEGY
 
 3.1 Two classical proofs of the Riemannian case, and one gap
 
-When K = 0 the Penrose inequality reads E >= sqrt(A/(16 pi)) in 3D,
+When K = 0 the Penrose ineuality reads E >= sqrt(A/(16 pi)) in 3D,
 E >= (1/2)(A/(2 pi^2))^(2/3) in 4D. Two complete proofs exist in 3D
 for arbitrary horizon area A (not just connected horizons):
 
@@ -495,7 +495,7 @@ for arbitrary horizon area A (not just connected horizons):
    Bray (2001), via a conformal flow of the metric.
 
 Both are deep and technically hard. In 4D, Bray-Lee (2009) proved the
-same inequality via the conformal flow approach; that proof works in
+same ineuality via the conformal flow approach; that proof works in
 dimensions up to seven.
 
 There is no analogous theorem in the general (K != 0) case. The
@@ -510,7 +510,7 @@ expresses the scalar curvature of a certain deformed metric in terms
 of original constraint densities, a nonnegative quadratic term, and
 a divergence. It suggested that if one can find a solution to a
 certain coupled system - the "generalized Jang equation" - then the
-spacetime Penrose inequality reduces to the Riemannian one.
+spacetime Penrose ineuality reduces to the Riemannian one.
 
 But: the coupled system that appeared in that approach was known
 only to have solutions in very special cases. Han-Khuri (2013)
@@ -526,7 +526,7 @@ depending strictly on graph height, and a hard penalty near the
 conformal lower bound - whose solutions do exist in general,
 uniformly across a large class of data. The result is
 a complete proof of a boundary version of the spacetime Penrose
-inequality, in both 3D and 4D.
+ineuality, in both 3D and 4D.
 
 
 3.2 The graph deformation idea
@@ -564,7 +564,7 @@ If both equations hold, the scalar curvature of ghat is exactly the
 quantity on the right of the identity. The trick is to choose F
 and Xi so that this quantity is nonnegative (which gives R_ghat
 >= 0), and so that the boundary conditions of the problem match
-those needed to apply the Riemannian Penrose inequality.
+those needed to apply the Riemannian Penrose ineuality.
 
 The construction then becomes a nonlinear elliptic system. The
 existence is proved via a continuation/degree argument: deform the
@@ -578,7 +578,7 @@ from the trivial endpoint.
 
 3.3 Why an inner boundary is useful
 
-The usual plan for the Penrose inequality is: "collapse" the
+The usual plan for the Penrose ineuality is: "collapse" the
 horizon, replacing the data inside it by something simpler. Close
 up the inner boundary into a smooth cap, work on a manifold without
 boundary, apply known results. This is the "filling" approach and it
@@ -659,8 +659,8 @@ maximal-vacuum exterior that is topologically a ball exterior with
 a torus symmetry. The extra hypothesis ensures positivity of a
 certain conformal factor that strictifies the dominant-energy
 condition, and the analysis of the coupled system is similar modulo
-the drift. The final inequality is the four-dimensional Riemannian
-Penrose inequality, applied to the deformed metric.
+the drift. The final ineuality is the four-dimensional Riemannian
+Penrose ineuality, applied to the deformed metric.
 
 For the general (non-maximal, non-compact-support) case, the paper
 uses the three-dimensional or four-dimensional boundary method
@@ -830,10 +830,10 @@ this document mostly speaks in 3D terms but the 4D case runs in
 parallel with these adjusted constants.
 
 
-4.3 Theorem C: the three-dimensional Penrose inequality
+4.3 Theorem C: the three-dimensional Penrose ineuality
 
-Combining Theorem A with the Riemannian Penrose inequality produces
-a proof of the three-dimensional spacetime Penrose inequality,
+Combining Theorem A with the Riemannian Penrose ineuality produces
+a proof of the three-dimensional spacetime Penrose ineuality,
 under the hypotheses of the prepared three-dimensional exterior.
 
 Specifically: let (Omega_0, g, K) be a prepared three-dimensional
@@ -848,7 +848,7 @@ S_0, computed in the metric g.
 This is stated as "Theorem 1.2" of the introduction. The key point:
 no extension of the data across the original boundary S_0 is
 required. Only the data on the exterior Omega_0 matters; the
-horizon is treated as an inequality, not as a surface you must cross.
+horizon is treated as an ineuality, not as a surface you must cross.
 
 Proof sketch:
 
@@ -863,7 +863,7 @@ Proof sketch:
       a compact minimal surface Gamma_N in Omega, detached from B,
       enclosing B.
 
-   3. Apply the Riemannian Penrose inequality of Bray (3D, full
+   3. Apply the Riemannian Penrose ineuality of Bray (3D, full
       perimeter, allowed to be disconnected) to the exterior of
       Gamma_N with metric ghat. This gives
 
@@ -906,7 +906,7 @@ Exactly analogous:
 
 This is stated as "Corollary 1.4" of the introduction and is the
 direct consequence of Theorem B plus the four-dimensional
-Riemannian Penrose inequality (Bray-Lee, 2009). The proof runs the
+Riemannian Penrose ineuality (Bray-Lee, 2009). The proof runs the
 same as the 3D case, with the four-dimensional rules:
 
    New metric ghat has R_ghat >= 0 and ghat >= e^{-2 eps} g.
@@ -917,7 +917,7 @@ same as the 3D case, with the four-dimensional rules:
 
    Combine, then limit N -> infinity, then eps -> 0.
 
-The result is the four-dimensional spacetime Penrose inequality in
+The result is the four-dimensional spacetime Penrose ineuality in
 the compact-support case and, via strictification, in the weakly
 decaying case.
 
@@ -925,7 +925,7 @@ decaying case.
 4.5 Theorem E: the four-dimensional maximal-vacuum case
 
 The final theorem states the "maximal-method" result: a sharp form
-of the Penrose inequality in four dimensions for a specific class
+of the Penrose ineuality in four dimensions for a specific class
 of maximal, vacuum, torus-symmetric data.
 
 SETUP. Let (M, g, K) be a four-dimensional initial-data exterior:
@@ -1993,7 +1993,7 @@ limiting value coming from the boundary expansion, allowing
 signed slopes. In the incompatible direction the correctors strictly
 violate this to force the slope back.
 
-Precise statement of the limiting directional inequality:
+Precise statement of the limiting directional ineuality:
 
    At any face point x with assigned height b, and considering the
    LIMITING case where |grad f| -> 1 (i.e., a = 1, chi = d = 0),
@@ -2091,7 +2091,7 @@ Step 1: A Gauss-type comparison at the minimum.
 
 At any interior minimum of t, we have grad t = 0 and Hess t >= 0.
 Under these constraints, the exact Gauss equation for the graph
-metric ghat reduces to a clean inequality:
+metric ghat reduces to a clean ineuality:
 
    (1/2) e^{4 t} R_ghat  <=  (1/2) R_g  -  v Ric_g(e, e)
                               + S(H^f),
@@ -2128,7 +2128,7 @@ c_* > 0 and a POLYNOMIAL (in N) multiplicative constant.
 The "v = 1 - d" in the second line is important: when v is small
 (near-graph-zero slope), the correction is small.
 
-The proof uses the Cauchy-Schwarz inequality |Fa q| <= chi q^2/2 +
+The proof uses the Cauchy-Schwarz ineuality |Fa q| <= chi q^2/2 +
 d^2 F^2/(2 chi), together with the denominator bound
 chi <= (n-1) d / (n-1 + pv) and the fact that p <= N.
 
@@ -2175,7 +2175,7 @@ Rearranging,
 
 Choose delta_0 < min(c_3/2, 1/4). Choose the penalty coefficient
 C_N so large (as a polynomial in N) that the (C_N - C_N) term
-absorbs the -rho term. Then the inequality is IMPOSSIBLE: LHS is
+absorbs the -rho term. Then the ineuality is IMPOSSIBLE: LHS is
 strictly negative, but RHS (the difference of the two sides of
 the divergence equation) is required to be zero.