[mcshicks]

You can run the model for free on openrouter.ai. I have played around with lean, slowly working my way through mathmatics in lean. I was stuck on this problem in section 3.6

theorem convergesTo_unique {s : ℕ → ℝ} {a b : ℝ} (sa : ConvergesTo s a) (sb : ConvergesTo s b) :

For fun I tried it on the free model on openrouter.ai. Got the answer the first time.

https://leanprover-community.github.io/mathematics_in_lean/m...

Here's the answer just to give you a feel.

  by_contra h
  have h₁ : a ≠ b := h
  have h₂ : |a - b| > 0 := by
    apply abs_pos.mpr
    exact sub_ne_zero.mpr h₁
  -- Use the definition of convergence to find N₁ and N₂
  have h₃ := sa (|a - b| / 2) (by linarith)
  have h₄ := sb (|a - b| / 2) (by linarith)
  cases' h₃ with N₁ h₃
  cases' h₄ with N₂ h₄
  -- Choose N to be the maximum of N₁ and N₂
  let N := max N₁ N₂
  have h₅ := h₃ N (by simp [N, le_max_left])
  have h₆ := h₄ N (by simp [N, le_max_right])
  -- Derive a contradiction using the triangle inequality
  have h₇ : |s N - a| < |a - b| / 2 := by simpa using h₅
  have h₈ : |s N - b| < |a - b| / 2 := by simpa using h₆
  have h₉ : |a - b| < |a - b| := by
    calc
      |a - b| = |a - s N + (s N - b)| := by ring_nf
      _ ≤ |a - s N| + |s N - b| := by
        apply abs_add
      _ = |s N - a| + |s N - b| := by
        rw [abs_sub_comm]
      _ < |a - b| / 2 + |a - b| / 2 := by
        linarith
      _ = |a - b| := by ring
  linarith