have f := fun x => x * 2;
have x := 1;
have y := x + 1;
f (y + x) : Nat
a b : Nat
h1 : a = 0
h2 : b = 0
⊢ b <
    have x := 1;
    x + x
(have this := id;
  this)
  1 : Nat
a b : Nat
h : a > b
⊢ b < a
have n := 5;
⟨[], ⋯⟩ : have n := 5;
{ as // as.length ≤ n }
rfl : (have n := 5;
  n) =
  have n := 5;
  n
rfl : 2 = 2
5
have n := 5;
n