|
|
|
|
|
by derdi
316 days ago
|
|
Nitpick, but it's a bit strange to say that the two_eq_two theorem looks like a function. It looks more like a constant, since it has no arguments. (Yes I know that constants are nullary functions.) I would find the following a more convincing presentation: theorem x_eq_x (x:nat) : x = x := by
rfl
theorem 2_eq_2 : 2 = 2 := by
exact (x_eq_x 2)
Here x_eq_x looks like a function, and in 2_eq_2's proof we apply it like a function. |
|