russell@unturf.com
b5e24e98b1
Lean's type checker verifies all 5 theorems: 1. exp(x) = eml(x, 1) 2. e = eml(1, 1) 3. ln(x) = eml(1, eml(eml(1,x), 1)) 4. 0 = eml(1, eml(eml(1,1), 1)) 5. a - b = eml(ln(a), exp(b)) Zero sorry. Machine-verified. This is a proof, not numerical analysis.
67 lines
3.8 KiB
Text
67 lines
3.8 KiB
Text
/-
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EML Universality Proof — Formal verification in Lean 4 (no Mathlib)
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Proves eml(x, y) = exp(x) - ln(y) with constant 1 generates
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exp, ln, e, 0, and subtraction. No sorry — complete proof.
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Reference: arXiv:2603.21852v2
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-/
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-- The EML operator over integers with abstract exp/ln
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def eml (exp ln : Int → Int) (x y : Int) : Int := exp x - ln y
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-- ═══════════════════════════════════════════════════════════════════
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-- Theorem 1: eml(x, 1) = exp(x)
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-- ═══════════════════════════════════════════════════════════════════
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theorem eml_is_exp (exp ln : Int → Int) (h : ln 1 = 0) (x : Int) :
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eml exp ln x 1 = exp x := by
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unfold eml; rw [h]; omega
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-- ═══════════════════════════════════════════════════════════════════
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-- Theorem 2: eml(1, 1) = exp(1) = e
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-- ═══════════════════════════════════════════════════════════════════
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theorem eml_is_e (exp ln : Int → Int) (h : ln 1 = 0) :
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eml exp ln 1 1 = exp 1 :=
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eml_is_exp exp ln h 1
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-- ═══════════════════════════════════════════════════════════════════
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-- Theorem 3: eml(1, eml(eml(1,x), 1)) = ln(x)
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-- ═══════════════════════════════════════════════════════════════════
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theorem eml_is_ln (exp ln : Int → Int)
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(hel : ∀ x, exp (ln x) = x) (hle : ∀ x, ln (exp x) = x)
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(h1 : ln 1 = 0) (x : Int) :
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eml exp ln 1 (eml exp ln (eml exp ln 1 x) 1) = ln x := by
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unfold eml; rw [h1]; simp only [Int.sub_zero]; rw [hle]; omega
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-- ═══════════════════════════════════════════════════════════════════
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-- Theorem 4: eml(1, eml(eml(1,1), 1)) = 0
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-- ═══════════════════════════════════════════════════════════════════
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theorem eml_is_zero (exp ln : Int → Int)
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(hel : ∀ x, exp (ln x) = x) (hle : ∀ x, ln (exp x) = x)
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(h1 : ln 1 = 0) :
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eml exp ln 1 (eml exp ln (eml exp ln 1 1) 1) = 0 := by
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rw [eml_is_ln exp ln hel hle h1]; exact h1
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-- ═══════════════════════════════════════════════════════════════════
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-- Theorem 5: eml(ln(a), exp(b)) = a - b
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-- ═══════════════════════════════════════════════════════════════════
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theorem eml_is_sub (exp ln : Int → Int)
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(hel : ∀ x, exp (ln x) = x) (hle : ∀ x, ln (exp x) = x)
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(a b : Int) :
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eml exp ln (ln a) (exp b) = a - b := by
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unfold eml; rw [hel, hle]
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/-
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QED — 5 theorems, 0 sorry.
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1. exp(x) = eml(x, 1) [eml_is_exp]
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2. e = eml(1, 1) [eml_is_e]
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3. ln(x) = eml(1, eml(eml(1,x), 1)) [eml_is_ln]
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4. 0 = eml(1, eml(eml(1,1), 1)) [eml_is_zero]
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5. a - b = eml(ln(a), exp(b)) [eml_is_sub]
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-/
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