diff --git a/examples/algebra.pg b/examples/algebra.pg
index 6ff2f33..0f02dd7 100644
--- a/examples/algebra.pg
+++ b/examples/algebra.pg
@@ -25,7 +25,6 @@ section BasicDefinitions
 
     -- fix some binary operation `op`
     variable (* : binop);
-
     infixl 20 *;
 
     -- it is associative if ...
diff --git a/examples/peano.pg b/examples/peano.pg
index f186953..debac3e 100644
--- a/examples/peano.pg
+++ b/examples/peano.pg
@@ -79,11 +79,15 @@ def three : nat := suc two;
 def four  : nat := suc three;
 def five  : nat := suc four;
 
+-- A nice infix abbreviation for equality on natural numbers
+def = (n m : nat) := eq nat n m;
+infixl 11 =;
+
 -- First, the predecessor of n is m if n = suc m.
-def pred (n m : nat) : ★ := eq nat n (suc m);
+def pred (n m : nat) : ★ := n = suc m;
 
 -- Our claim is a disjunction, whose first option is that n = 0.
-def szc_l (n : nat) := eq nat n zero;
+def szc_l (n : nat) := n = zero;
 
 -- The second option is that n has a predecessor.
 def szc_r (n : nat) := exists nat (pred n);
@@ -194,7 +198,7 @@ def rec_rel_total : total nat A rec_rel :=
 
 -- {{{ Defining R2
 def alt_cond1 (x : nat) (y : A) :=
-    eq nat x zero ∧ eq A y z;
+    x = zero ∧ eq A y z;
 
 def cond_y2 (x : nat) (y : A)
     (x2 : nat) (y2 : A) :=
@@ -213,8 +217,8 @@ def rec_rel_alt (x : nat) (y : A) := alt_cond1 x y ∨ alt_cond2 x y;
 -- {{{ R2 ⊆ R
 def R2_sub_R_case1 (x : nat) (y : A) : alt_cond1 x y -> rec_rel x y :=
     fun (case1 : alt_cond1 x y) =>
-        let (x0 := and_elim_l (eq nat x zero) (eq A y z) case1)
-            (yz := and_elim_r (eq nat x zero) (eq A y z) case1)
+        let (x0 := and_elim_l (x = zero) (eq A y z) case1)
+            (yz := and_elim_r (x = zero) (eq A y z) case1)
             (a1 := (eq_sym A y z yz) (rec_rel zero) rec_rel_cond1)
         in
             (eq_sym nat x zero x0) (fun (n : nat) => rec_rel n y) a1
@@ -252,7 +256,7 @@ def R2_sub_R (x : nat) (y : A) : rec_rel_alt x y -> rec_rel x y :=
 -- {{{ R ⊆ R2
 def R2_cond1 : cond1 rec_rel_alt :=
     or_intro_l (alt_cond1 zero z) (alt_cond2 zero z)
-    (and_intro (eq nat zero zero) (eq A z z)
+    (and_intro (zero = zero) (eq A z z)
         (eq_refl nat zero)
         (eq_refl A z));
 
@@ -292,7 +296,7 @@ def R2_zero (y : A) : rec_rel_alt zero y -> eq A y z :=
         (cy2 := cond_y2 zero y)
     in fun (h : rec_rel_alt zero y) =>
         or_elim (alt_cond1 zero y) (alt_cond2 zero y) (eq A y z) h
-        (fun (case1 : alt_cond1 zero y) => and_elim_r (eq nat zero zero) (eq A y z) case1)
+        (fun (case1 : alt_cond1 zero y) => and_elim_r (zero = zero) (eq A y z) case1)
         (fun (case2 : alt_cond2 zero y) =>
             exists_elim nat (eq A y z) cx2 case2
             (fun (x2 : nat) (h2 : cx2 x2) =>
@@ -309,7 +313,7 @@ def R2_suc (x2 : nat) (y : A) : rec_rel_alt (suc x2) y -> exists A (cond_y2 (suc
     in
         fun (h : rec_rel_alt x y) =>
         or_elim (alt_cond1 x y) (alt_cond2 x y) goal h
-        (fun (case1 : alt_cond1 x y) => suc_nonzero x2 (and_elim_l (eq nat x zero) (eq A y z) case1) goal)
+        (fun (case1 : alt_cond1 x y) => suc_nonzero x2 (and_elim_l (x = zero) (eq A y z) case1) goal)
         (fun (case2 : alt_cond2 x y) =>
             exists_elim nat goal cx2 case2
             (fun (x22 : nat) (hx22 : cx2 x22) =>
@@ -446,17 +450,17 @@ infixl 20 +;
 
 -- The first equation manifests itself as the familiar
 --      n + 0 = 0.
-def plus_0_r (n : nat) : eq nat (n + zero) n :=
+def plus_0_r (n : nat) : n + zero = n :=
   rec_def_sat_zero nat n psuc;
 
 -- The second equation, after a bit of massaging, manifests itself as the
 -- likewise familiar
 --      n + suc m = suc (n + m).
-def plus_s_r (n m : nat) : eq nat (n + suc m) (suc (n + m)) :=
+def plus_s_r (n m : nat) : n + suc m = suc (n + m) :=
   rec_def_sat_suc nat n psuc m;
 
 -- -- We can now prove 1 + 1 = 2!
-def one_plus_one_two : eq nat (one + one) two :=
+def one_plus_one_two : one + one = two :=
   -- 1 + (suc zero) = suc (1 + zero) = suc one
   eq_trans nat (one + one) (suc (one + zero)) two
 
@@ -473,11 +477,11 @@ def one_plus_one_two : eq nat (one + one) two :=
 -- We will now prove a couple standard properties of addition.
 
 -- First, associativity, namely that n + (m + p) = (n + m) + p.
-def plus_assoc : forall (n m p : nat), eq nat (n + (m + p)) (n + m + p)
+def plus_assoc : forall (n m p : nat), n + (m + p) = n + m + p
   := fun (n m : nat) =>
   -- We prove this via induction on p for any fixed n and m.
   nat_ind
-    (fun (p : nat) => eq nat (n + (m + p)) (n + m + p))
+    (fun (p : nat) => n + (m + p) = n + m + p)
 
     -- Base case: p = 0
     -- WTS n + (m + 0) = (n + m) + 0
@@ -490,7 +494,7 @@ def plus_assoc : forall (n m p : nat), eq nat (n + (m + p)) (n + m + p)
       (eq_sym nat (n + m + zero) (n + m) (plus_0_r (n + m))))
 
     -- Inductive step: IH = n + (m + p) = (n + m) + p
-    (fun (p : nat) (IH : eq nat (n + (m + p)) (n + m + p)) =>
+    (fun (p : nat) (IH : n + (m + p) = n + m + p) =>
       -- WTS n + (m + suc p) = (n + m) + suc p
       -- n + (m + suc p) = n + suc (m + p)
       --                 = suc (n + (m + p))
@@ -518,32 +522,32 @@ def plus_assoc : forall (n m p : nat), eq nat (n + (m + p)) (n + m + p)
 -- Up next is commutativity, but we will need a couple lemmas first.
 
 -- First, we will show that 0 + n = n.
-def plus_0_l : forall (n : nat), eq nat (zero + n) n :=
+def plus_0_l : forall (n : nat), zero + n = n :=
   -- We prove this by induction on n.
-  nat_ind (fun (n : nat) => eq nat (zero + n) n)
+  nat_ind (fun (n : nat) => zero + n = n)
     -- base case: WTS 0 + 0 = 0
     -- This is just plus_0_r 0
     (plus_0_r zero)
 
     -- inductive case
-    (fun (n : nat) (IH : eq nat (zero + n) n) =>
+    (fun (n : nat) (IH : zero + n = n) =>
       -- WTS 0 + (suc n) = suc n
-      -- 0 + (suc n) = suc (0 + n) = suc n
+      -- 0 + suc n = suc (0 + n) = suc n
       eq_trans nat (zero + suc n) (suc (zero + n)) (suc n)
-      -- 0 + (suc n) = suc (0 + n)
+      -- 0 + suc n = suc (0 + n)
       (plus_s_r zero n)
 
       -- suc (0 + n) = suc n
       (eq_cong nat nat (zero + n) n suc IH));
 
 -- Next, we will show that (suc n) + m = suc (n + m).
-def plus_s_l (n : nat) : forall (m : nat), eq nat (suc n + m) (suc (n + m)) :=
+def plus_s_l (n : nat) : forall (m : nat), suc n + m = suc (n + m) :=
   -- We proceed by induction on m.
-  nat_ind (fun (m : nat) => eq nat (suc n + m) (suc (n + m)))
+  nat_ind (fun (m : nat) => suc n + m = suc (n + m))
     -- base case: (suc n) + 0 = suc (n + 0)
-    -- (suc n) + 0 = suc n = suc (n + 0)
+    -- suc n + 0 = suc n = suc (n + 0)
     (eq_trans nat (suc n + zero) (suc n) (suc (n + zero))
-      -- (suc n) + 0 = suc n
+      -- suc n + 0 = suc n
       (plus_0_r (suc n))
 
       -- suc n = suc (n + 0)
@@ -553,7 +557,7 @@ def plus_s_l (n : nat) : forall (m : nat), eq nat (suc n + m) (suc (n + m)) :=
 
     -- inductive case
     -- IH = suc n + m = suc (n + m)
-    (fun (m : nat) (IH : eq nat (suc n + m) (suc (n + m))) =>
+    (fun (m : nat) (IH : suc n + m = suc (n + m)) =>
         -- WTS suc n + suc m = suc (n + suc m)
         -- suc n + suc m = suc (suc n + m)
         --               = suc (suc (n + m))
@@ -573,9 +577,9 @@ def plus_s_l (n : nat) : forall (m : nat), eq nat (suc n + m) (suc (n + m)) :=
               (eq_sym nat (n + suc m) (suc (n + m)) (plus_s_r n m))))));
 
 -- Finally, we can prove commutativity.
-def plus_comm (n : nat) : forall (m : nat), eq nat (n + m) (m + n) :=
+def plus_comm (n : nat) : forall (m : nat), n + m = m + n :=
   -- As usual, we proceed by induction.
-  nat_ind (fun (m : nat) => eq nat (n + m) (m + n))
+  nat_ind (fun (m : nat) => n + m = m + n)
 
   -- Base case: WTS n + 0 = 0 + n
   -- n + 0 = n = 0 + n
@@ -587,7 +591,7 @@ def plus_comm (n : nat) : forall (m : nat), eq nat (n + m) (m + n) :=
     (eq_sym nat (zero + n) n (plus_0_l n)))
 
   -- Inductive step:
-  (fun (m : nat) (IH : eq nat (n + m) (m + n)) =>
+  (fun (m : nat) (IH : n + m = m + n) =>
     -- WTS n + suc m = suc m + n
     -- n + suc m = suc (n + m)
     --           = suc (m + n)