diff --git a/three_manifold_no_conjugate_v2.txt b/three_manifold_no_conjugate_v2.txt
index 034ebb9..3dd383c 100644
--- a/three_manifold_no_conjugate_v2.txt
+++ b/three_manifold_no_conjugate_v2.txt
@@ -699,7 +699,7 @@ in BOTH end frames T_1, T_2:
 
   (c)  -P''(t)  >=  0         (symmetric positive-semidefinite);
 
-  (d)  |P'(t) p|^2  <=  C δ \, p^T (-P''(t)) p    for every p ∈ R^2.
+  (d)  |P'(t) p|^2  <=  C δ p^T (-P''(t)) p    for every p ∈ R^2.
 
 Here ||.|| is the operator norm on symmetric 2x2 matrices.  The
 constants δ_0 and C are independent of B_1 and B_2.
@@ -873,7 +873,7 @@ Bound (d) on I_0: for p ∈ R^2,
     |P'p|^2 <= ||P'||^2 |p|^2 <= (1+C_1)^2 δ^2 |p|^2,
 and by (6.17)   p^T(-P'')p  >=  (δ m / 2) |p|^2.  Hence
 
-    |P'p|^2  <=  ( 2 (1+C_1)^2 / m ) δ \, p^T(-P'')p.
+    |P'p|^2  <=  ( 2 (1+C_1)^2 / m ) δ p^T(-P'')p.
 
 Take C := max( 3, 1 + C_1, 2 (1+C_1)^2 / m ).  Then (b), (c), (d) hold
 in the original frame.
@@ -891,7 +891,7 @@ For the domination, set v = T_2^{-T} p so that  p^T(-P_2'')p = v^T(-P'')v:
     |P_2' p|^2  =  |T_2^{-1} P'v|^2
                 <= ||T_2^{-1}||^2 |P'v|^2
                 <= ||T_2^{-1}||^2 C δ v^T(-P'')v
-                <= 3 C δ \, p^T(-P_2'')p.
+                <= 3 C δ p^T(-P_2'')p.
 
 Replacing C by 3 C finishes the proof of Lemma 6.4.
 
@@ -1150,7 +1150,7 @@ continuous.
 Summary of Part 6.  We have built a C^∞ Riemannian metric g on the
 closed, connected, orientable 3-manifold M, with the property
 
-    sec(g) <= 0    on    M  \  ( (0,1) x T^2 )                 (6.33)
+    sec(g) <= 0    on    M  minus  ( (0,1) x T^2 )                 (6.33)
 
 (the open neck is the only place where positive curvature can occur).
 In addition sec(g) <= 0 on the two end portions t ∈ (0, 1/4] and
@@ -2522,7 +2522,7 @@ neck).  We must now sum over passages.
 
 PROPOSITION 10.2 (Assembly of passage estimates).  Let (M, g) be a
 complete smooth Riemannian manifold, U ⊂ M open.  Suppose:
-  (i) sec_M <= 0 on M \ U;  and
+  (i) sec_M <= 0 on M minus U;  and
   (ii) for every complete unit-speed geodesic gamma: R -> M, every
        connected component J of gamma^{-1}(U), and every smooth compactly
        supported field V along gamma, we have I_J(V) >= 0.
@@ -2531,7 +2531,7 @@ Then g has no conjugate points.
 Proof.  Fix gamma and V.  The set gamma^{-1}(U) ⊂ R is open, so it is
 a COUNTABLE disjoint union of open intervals J_j (each contains a
 distinct rational number, so there are only countably many).  The
-complement E := R \ gamma^{-1}(U) is closed.
+complement E := R minus gamma^{-1}(U) is closed.
 
 On each J_j, hypothesis (ii) gives I_{J_j}(V) >= 0.