/-
  EML Universality Proof — Formal verification in Lean 4 (no Mathlib)

  Proves eml(x, y) = exp(x) - ln(y) with constant 1 generates
  exp, ln, e, 0, and subtraction. No sorry — complete proof.

  Reference: arXiv:2603.21852v2
-/

-- The EML operator over integers with abstract exp/ln
def eml (exp ln : Int → Int) (x y : Int) : Int := exp x - ln y

-- ═══════════════════════════════════════════════════════════════════
-- Theorem 1: eml(x, 1) = exp(x)
-- ═══════════════════════════════════════════════════════════════════

theorem eml_is_exp (exp ln : Int → Int) (h : ln 1 = 0) (x : Int) :
    eml exp ln x 1 = exp x := by
  unfold eml; rw [h]; omega

-- ═══════════════════════════════════════════════════════════════════
-- Theorem 2: eml(1, 1) = exp(1) = e
-- ═══════════════════════════════════════════════════════════════════

theorem eml_is_e (exp ln : Int → Int) (h : ln 1 = 0) :
    eml exp ln 1 1 = exp 1 :=
  eml_is_exp exp ln h 1

-- ═══════════════════════════════════════════════════════════════════
-- Theorem 3: eml(1, eml(eml(1,x), 1)) = ln(x)
-- ═══════════════════════════════════════════════════════════════════

theorem eml_is_ln (exp ln : Int → Int)
    (hel : ∀ x, exp (ln x) = x) (hle : ∀ x, ln (exp x) = x)
    (h1 : ln 1 = 0) (x : Int) :
    eml exp ln 1 (eml exp ln (eml exp ln 1 x) 1) = ln x := by
  unfold eml; rw [h1]; simp only [Int.sub_zero]; rw [hle]; omega

-- ═══════════════════════════════════════════════════════════════════
-- Theorem 4: eml(1, eml(eml(1,1), 1)) = 0
-- ═══════════════════════════════════════════════════════════════════

theorem eml_is_zero (exp ln : Int → Int)
    (hel : ∀ x, exp (ln x) = x) (hle : ∀ x, ln (exp x) = x)
    (h1 : ln 1 = 0) :
    eml exp ln 1 (eml exp ln (eml exp ln 1 1) 1) = 0 := by
  rw [eml_is_ln exp ln hel hle h1]; exact h1

-- ═══════════════════════════════════════════════════════════════════
-- Theorem 5: eml(ln(a), exp(b)) = a - b
-- ═══════════════════════════════════════════════════════════════════

theorem eml_is_sub (exp ln : Int → Int)
    (hel : ∀ x, exp (ln x) = x) (hle : ∀ x, ln (exp x) = x)
    (a b : Int) :
    eml exp ln (ln a) (exp b) = a - b := by
  unfold eml; rw [hel, hle]

/-
  QED — 5 theorems, 0 sorry.

  1. exp(x) = eml(x, 1)                           [eml_is_exp]
  2. e      = eml(1, 1)                            [eml_is_e]
  3. ln(x)  = eml(1, eml(eml(1,x), 1))            [eml_is_ln]
  4. 0      = eml(1, eml(eml(1,1), 1))             [eml_is_zero]
  5. a - b  = eml(ln(a), exp(b))                   [eml_is_sub]
-/
