diff --git a/markov_type_superreflexivity_v2.txt b/markov_type_superreflexivity_v2.txt
index 64d3ad8..9f7ad76 100644
--- a/markov_type_superreflexivity_v2.txt
+++ b/markov_type_superreflexivity_v2.txt
@@ -1,4 +1,4 @@
-The Markov type inequality tests every finite reversible random walk at
+The Markov type ineuality tests every finite reversible random walk at
 once. Let S be a finite set, P a transition matrix on S with stationary
 law pi satisfying pi_i P_ij = pi_j P_ji for all i,j in S, and write
 Z_0, Z_1, Z_2, ... for the chain started in pi. A real Banach space X has
@@ -8,7 +8,7 @@ every map f from S to X and every integer t >= 1,
      E ||f(Z_t) - f(Z_0)||^p <= K^p t E ||f(Z_1) - f(Z_0)||^p.
 
 The left side measures how far a walk of length t travels, the right side
-measures one step. The inequality says that the p-th moment of the
+measures one step. The ineuality says that the p-th moment of the
 displacement grows at most linearly in time, with one constant K serving
 all walks and all targets. Keith Ball introduced the condition to prove
 Lipschitz extension theorems, and the only structure it sees is a
@@ -42,10 +42,10 @@ norms. Hence a real Banach space is superreflexive exactly when it has
 Markov type p for some p > 1, which answers a question of Assaf Naor:
 whether nontrivial Markov type can exist without an equivalent uniformly
 smooth norm.
-https://arxiv.org/pdf/math/0410422
+./pdf/unknown-nd-math0410422.pdf
 ./pdf/npss-2006.pdf
 https://doi.org/10.1007/BF02760337
-https://arxiv.org/pdf/1205.5993
+./pdf/unknown-nd-1205.5993.pdf
 ./pdf/naor-ribe-2012.pdf
 
 Rademacher type, the weaker cousin, does not suffice. James constructed a
@@ -66,7 +66,7 @@ https://doi.org/10.1007/BF02385837
 https://doi.org/10.1007/BF02766125
 https://doi.org/10.1142/S1793525309000114
 https://doi.org/10.1007/s00039-008-0689-0
-https://arxiv.org/pdf/0803.1697
+./pdf/unknown-nd-0803.1697.pdf
 ./pdf/mendel-naor-2013.pdf
 
 The proof runs by contradiction in four moves, and only four facts are
@@ -162,7 +162,7 @@ y_{i_{j+1}},...,y_{i_m}. Hence f_{i_{j+1}}(t) lies within
      ||t|| >= d max_{0<=j<m} |a_{j+1} + ... + a_m|
                 - (1/M) sum_h |a_h|,
 
-while the triangle inequality and ||y|| <= 1 give ||t|| <= sum_h |a_h|.
+while the triangle ineuality and ||y|| <= 1 give ||t|| <= sum_h |a_h|.
 Read from the far end, where a_1 sits on the largest of the chosen
 vectors, the lower bound controls the prefix sums of the list a.
 
@@ -228,7 +228,7 @@ most N_0(F_J a). In block i of the longer list choose a uniformly random
 subset of size J_i, put a_i/J_i on that subset and zeros elsewhere, and do
 this independently in each block. Every outcome is F_J a with extra zeros
 inserted, so every outcome has N_0-norm N_0(F_J a), the mean of the
-outcomes is F_M a, and convexity of N_0 gives the inequality.
+outcomes is F_M a, and convexity of N_0 gives the ineuality.
 
 The numbers N_0(F_J a) decrease as J grows and stay bounded below, since
 the prefix sums of a survive at block ends, so the limit
@@ -317,7 +317,7 @@ Then for every n >= 1,
 
      E || g(E_1) + ... + g(E_n) ||^p <= K^p n E || g(E_1) ||^p.
 
-Proof. The inequality is invariant under scaling, and if g vanishes
+Proof. The ineuality is invariant under scaling, and if g vanishes
 identically there is nothing to prove. Otherwise choose one edge e_j from
 each two-element orbit of the involution, j = 1,...,r, and let u_1,...,u_r
 be the standard basis of Z^r. Give e_j the displacement u_j, give e_j^* the
@@ -361,7 +361,7 @@ chain, gives
        <= E || Phi(Z_n) - Phi(Z_0) ||^p
        <= K^p n E || g(E_1) ||^p.
 
-Let R tend to infinity. The same inequality holds for the target X by
+Let R tend to infinity. The same ineuality holds for the target X by
 applying Markov type there to the composition of Phi with T, and then
 letting eta tend to 0.
 
@@ -629,7 +629,7 @@ Put L = 8(n+1) and eps = 1/(10n), and for k >= 1 put
      D_k = N(1,-1,1,-1,...,1,-1)      with k pairs.
 
 Each D_k is positive, since N is a norm and the list is nonzero, and
-spreading invariance together with the triangle inequality gives
+spreading invariance together with the triangle ineuality gives
 D_{k+l} <= D_k + D_l, so D_L <= 2 D_{L/2} and N of a run of k pairs
 (+c,-c) equals |c| D_k.
 
@@ -748,7 +748,7 @@ Take a = L/4 and b = 3L/4 and list the cuts in increasing order:
 
      0, r_a, t_a, r_{a+1}, t_{a+1}, ..., r_{b-1}, t_{b-1}, r_b, B.
 
-All of them are distinct. The inequalities r_j < t_j < r_{j+1} come from
+All of them are distinct. The ineualities r_j < t_j < r_{j+1} come from
 the interlacing, the two ends are strict because C_0(a) >= 1 and because
 the point L + 2n of Lambda is larger than b, so C_0(b) is not the largest
 rank. With F(0) = F(B) = 0 the prefix sums at the cuts, and the sums of G
@@ -767,7 +767,7 @@ applied inside the blocks, so
 
 Write that array as the sum of A = (V,0,...,0,-V) and
 B' = (0,ell,-ell,...,ell,-ell,0). Then N(A) = |V| D_1 and N(B') =
-ell D_{L/2}, so the triangle inequality in the form
+ell D_{L/2}, so the triangle ineuality in the form
 N(B') <= N(G) + N(A) gives
 
      N(G) >= ell D_{L/2} - |V| D_1 >= ell (D_{L/2} - D_1),
@@ -840,7 +840,7 @@ Since p > 1 the left side tends to infinity with n while the right side is
 fixed. This is impossible, so X is superreflexive.
 
 Three remarks on what entered where. The hypothesis p > 1 is used exactly
-once, in the last display; for p = 1 the inequalities are consistent, and
+once, in the last display; for p = 1 the ineualities are consistent, and
 indeed no contradiction is available. The finiteness of K is used once,
 when the odd-label lemma is applied. Everything else in the argument is
 quantitative in n but not in K: the walk constructed for a given n has