← Session 29 — ii′-5e-v: circle bound (far-tail + series bounds)
thinking (3857 chars)
Error at 1973 — h2's proof: `rw [div_le_div_iff₀ ...]` failed because div_le_div_iff₀ produces an iff, and rw with an iff rewrites the GOAL — the error shows the pattern mismatch: `div_le_div_iff₀` has a different form in this Mathlib. The goal: `R / (‖ρ‖ * (R−‖ρ‖)) ≤ 2 / (R − ‖ρ‖)`. div_le_div_iff₀ (hb : 0 < b) (hd : 0 < d) : a/b ≤ c/d ↔ a*d ≤ c*b. rw [div_le_div_iff₀ h1 h2] should turn the goal into R·(R−‖ρ‖) ≤ 2·(‖ρ‖·(R−‖ρ‖)). The error says the pattern with (‖ρ‖ * (‖ρ‖ * ...)) didn't match — the div_le_div_iff₀ I applied had the denominators unified wrongly: my first arg was `mul_pos hρpos ...` : 0 < ‖ρ‖*(R−‖ρ‖) ✓ b := ‖ρ‖*(R−‖ρ‖); second : 0 < R−‖ρ‖ ✓ d := R−‖ρ‖. Then the iff: a/b ≤ c/d ↔ a·d ≤ c·b with a := R, c := 2: R·(R−‖ρ‖) ≤ 2·(‖ρ‖·(R−‖ρ‖)). Hmm the error pattern shows `?m.1048 / (‖↑ρ‖ * (‖↑ρ‖ * (R - ‖↑ρ‖)))` — the b unified as ‖ρ‖·(‖ρ‖·(R−‖ρ‖))?? That means div_le_div_iff₀'s FIRST hypothesis is 0 < b where the goal's FIRST denominator is... The goal LHS: `R / (‖↑ρ‖ * (R - ‖↑ρ‖))` — the denominator is ‖ρ‖·(R−‖ρ‖). But the pattern shows ‖ρ‖ * (‖ρ‖ * (R−‖ρ‖)) — as if a/b parsed with b = ‖ρ‖·(‖ρ‖·(R−‖ρ‖))...
OH — I bet this Mathlib's div_le_div_iff₀ has signature `div_le_div_iff₀ (hb : 0 < b) (hd : 0 < d) : a / b ≤ c / d ↔ a * d ≤ c * b`? The rw failed to match the goal — maybe the name is different: div_le_div_iff_of_pos? or div_le_div_iff (no ₀)? The error shows the REWRITE tried to match `?a / ?b ≤ ?c / ?d` with ?b := ‖ρ‖·(‖ρ‖·(R−‖ρ‖)) — that's odd. Actually maybe div_le_div_iff₀ in this Mathlib is: `a / b ≤ c / d ↔ a * d ≤ c * b` requires matching the EXACT term `R / (‖ρ‖ * (R − ‖ρ‖))` — should match...
Wait, look at the pattern more carefully: `?m.1048 / (‖↑ρ‖ * (‖↑ρ‖ * (R - ‖↑ρ‖))) ≤ ?m.1050 / (R - ‖↑ρ‖)` — the LHS denominator is ‖ρ‖·(‖ρ‖·(R−‖ρ‖)) — that's ‖ρ‖ TIMES the intended denominator. This suggests div_le_div_iff₀'s form is different — perhaps it's `div_le_div_iff₀ : b / a ≤ d / c ↔ ...`? Or the lemma is stated as `(hb : 0 < b) (hd : 0 < d) : a / (b * c) ≤ ...`?
Hmm — in recent Mathlib, `div_le_div_iff₀` might be: (hb : 0 < b) (hd : 0 < d) : a / b ≤ c / d ↔ a * d ≤ c * b. Let me just avoid the rw and prove h2 directly:
h2 : R / (‖ρ‖·(R−‖ρ‖)) ≤ 2/(R−‖ρ‖):
```
have h2 : R / (‖(ρ : ℂ)‖ * (R - ‖(ρ : ℂ)‖)) ≤ 2 / (R - ‖(ρ : ℂ)‖) := by
rw [div_mul_eq_mul_div, div_le_iff₀ (by linarith [hd2] : (0:ℝ) < R - ‖(ρ:ℂ)‖)]
...
```
Alternative cleaner: multiply both sides: R/(‖ρ‖·d) ≤ 2/d ⟺ R·d ≤ 2·(‖ρ‖·d) (mult by d > 0... and divide by d): R/(‖ρ‖·d) = (R/‖ρ‖)/d: then (R/‖ρ‖)/d ≤ 2/d ⟸ R/‖ρ‖ ≤ 2 (div_le_div_of_nonneg_right with d > 0). R/‖ρ‖ ≤ 2 ⟺ R ≤ 2‖ρ‖ ✓ (h5). So:
```
have h2a : R / ‖(ρ : ℂ)‖ ≤ 2 := by
rw [div_le_iff₀ hρpos]
linarith [h5] -- R ≤ 2‖ρ‖: h5 : R ≤ 2·‖ρ‖ ✓... div_le_iff₀ (hb : 0 < b) : a / b ≤ c ↔ a ≤ c * b: R ≤ 2·‖ρ‖ ✓ h5
have h2 : R / (‖(ρ : ℂ)‖ * (R - ‖(ρ : ℂ)‖)) ≤ 2 / (R - ‖(ρ : ℂ)‖) := by
rw [show R / (‖(ρ : ℂ)‖ * (R - ‖(ρ : ℂ)‖)) = (R / ‖(ρ : ℂ)‖) / (R - ‖(ρ : ℂ)‖) from by rw [div_div]]
exact div_le_div_of_nonneg_right h2a (by linarith [hd2] : (0 : ℝ) < R - ‖(ρ : ℂ)‖)
```
Hmm — div_le_div_of_nonneg_right : a ≤ b → 0 < c → a/c ≤ b/c? The name in current Mathlib: `div_le_div_of_nonneg_right (hab : a ≤ b) (hc : 0 < c) : a / c ≤ b / c`? I recall `div_le_div_of_nonneg_right` hmm — there's div_le_div_right (ordered field): `div_le_div_right (hc : 0 < c) : a / c ≤ b / c ↔ a ≤ b`. So: `(div_le_div_right (by linarith [hd2] : (0:ℝ) < R - ‖(ρ:ℂ)‖)).mpr h2a` ✓ robust.
div_div : a / (b * c)?? div_div : a / b / c = a / (b * c) — so a/(b·c) = a/b/c via (div_div a b c).symm. My show: R/(‖ρ‖·d) = (R/‖ρ‖)/d: `rw [div_div]` on (R/‖ρ‖)/d gives R/(‖ρ‖·d) — so the show's proof is `by rw [div_div]` ✓.
Also check div_le_iff₀ exists (used in shell_low h3 — that compiled ✓ `div_le_iff₀ (by linarith [hR] : (0 : ℝ) < R - 3)` gave R ≤ 2·(R−3) ✓ worked).
Let me fix h2.
`div_le_div_iff₀`'s form mismatched. Rerouting h2 via `R/‖ρ‖ ≤ 2` + `div_le_div_right`.
tool: Edit
input
{
"replace_all": false,
"file_path": "D:\\Python\\Quantum\\research\\lean\\NguyenRH1\\NguyenRH1\\LiCircleBound.lean",
"old_string": " have h2 : R / (\u2016(\u03c1 : \u2102)\u2016 * (R - \u2016(\u03c1 : \u2102)\u2016)) \u2264 2 / (R - \u2016(\u03c1 : \u2102)\u2016) := by\n rw [div_le_div_iff\u2080 (mul_pos h\u03c1pos (by linarith [hd2] :\n (0 : \u211d) < \u2016(\u03c1 : \u2102)\u2016 * (R - \u2016(\u03c1 : \u2102)\u2016)))\n (by linarith [hd2] : (0 : \u211d) < R - \u2016(\u03c1 : \u2102)\u2016)]\n have h5 : R \u2264 2 * \u2016(\u03c1 : \u2102)\u2016 := by linarith [h\u03c1R2']\n have h6 : (0 : \u211d) \u2264 R - \u2016(\u03c1 : \u2102)\u2016 := by linarith [hd2]\n nlinarith [h5, h6, h\u03c1pos]",
"new_string": " have h2 : R / (\u2016(\u03c1 : \u2102)\u2016 * (R - \u2016(\u03c1 : \u2102)\u2016)) \u2264 2 / (R - \u2016(\u03c1 : \u2102)\u2016) := by\n have h5 : R \u2264 2 * \u2016(\u03c1 : \u2102)\u2016 := by linarith [h\u03c1R2']\n have h2a : R / \u2016(\u03c1 : \u2102)\u2016 \u2264 2 := by\n rw [div_le_iff\u2080 h\u03c1pos]\n linarith [h5]\n rw [show R / (\u2016(\u03c1 : \u2102)\u2016 * (R - \u2016(\u03c1 : \u2102)\u2016))\n = (R / \u2016(\u03c1 : \u2102)\u2016) / (R - \u2016(\u03c1 : \u2102)\u2016) from by rw [div_div]]\n exact (div_le_div_right (by linarith [hd2] : (0 : \u211d) < R - \u2016(\u03c1 : \u2102)\u2016)).mpr h2a"
}
raw content_json (database row)
[{"type":"thinking","thinking":"Error at 1973 — h2's proof: `rw [div_le_div_iff₀ ...]` failed because div_le_div_iff₀ produces an iff, and rw with an iff rewrites the GOAL — the error shows the pattern mismatch: `div_le_div_iff₀` has a different form in this Mathlib. The goal: `R / (‖ρ‖ * (R−‖ρ‖)) ≤ 2 / (R − ‖ρ‖)`. div_le_div_iff₀ (hb : 0 < b) (hd : 0 < d) : a/b ≤ c/d ↔ a*d ≤ c*b. rw [div_le_div_iff₀ h1 h2] should turn the goal into R·(R−‖ρ‖) ≤ 2·(‖ρ‖·(R−‖ρ‖)). The error says the pattern with (‖ρ‖ * (‖ρ‖ * ...)) didn't match — the div_le_div_iff₀ I applied had the denominators unified wrongly: my first arg was `mul_pos hρpos ...` : 0 < ‖ρ‖*(R−‖ρ‖) ✓ b := ‖ρ‖*(R−‖ρ‖); second : 0 < R−‖ρ‖ ✓ d := R−‖ρ‖. Then the iff: a/b ≤ c/d ↔ a·d ≤ c·b with a := R, c := 2: R·(R−‖ρ‖) ≤ 2·(‖ρ‖·(R−‖ρ‖)). Hmm the error pattern shows `?m.1048 / (‖↑ρ‖ * (‖↑ρ‖ * (R - ‖↑ρ‖)))` — the b unified as ‖ρ‖·(‖ρ‖·(R−‖ρ‖))?? That means div_le_div_iff₀'s FIRST hypothesis is 0 < b where the goal's FIRST denominator is... The goal LHS: `R / (‖↑ρ‖ * (R - ‖↑ρ‖))` — the denominator is ‖ρ‖·(R−‖ρ‖). But the pattern shows ‖ρ‖ * (‖ρ‖ * (R−‖ρ‖)) — as if a/b parsed with b = ‖ρ‖·(‖ρ‖·(R−‖ρ‖))... \n\nOH — I bet this Mathlib's div_le_div_iff₀ has signature `div_le_div_iff₀ (hb : 0 < b) (hd : 0 < d) : a / b ≤ c / d ↔ a * d ≤ c * b`? The rw failed to match the goal — maybe the name is different: div_le_div_iff_of_pos? or div_le_div_iff (no ₀)? The error shows the REWRITE tried to match `?a / ?b ≤ ?c / ?d` with ?b := ‖ρ‖·(‖ρ‖·(R−‖ρ‖)) — that's odd. Actually maybe div_le_div_iff₀ in this Mathlib is: `a / b ≤ c / d ↔ a * d ≤ c * b` requires matching the EXACT term `R / (‖ρ‖ * (R − ‖ρ‖))` — should match...\n\nWait, look at the pattern more carefully: `?m.1048 / (‖↑ρ‖ * (‖↑ρ‖ * (R - ‖↑ρ‖))) ≤ ?m.1050 / (R - ‖↑ρ‖)` — the LHS denominator is ‖ρ‖·(‖ρ‖·(R−‖ρ‖)) — that's ‖ρ‖ TIMES the intended denominator. This suggests div_le_div_iff₀'s form is different — perhaps it's `div_le_div_iff₀ : b / a ≤ d / c ↔ ...`? Or the lemma is stated as `(hb : 0 < b) (hd : 0 < d) : a / (b * c) ≤ ...`? \n\nHmm — in recent Mathlib, `div_le_div_iff₀` might be: (hb : 0 < b) (hd : 0 < d) : a / b ≤ c / d ↔ a * d ≤ c * b. Let me just avoid the rw and prove h2 directly:\n\nh2 : R / (‖ρ‖·(R−‖ρ‖)) ≤ 2/(R−‖ρ‖): \n```\nhave h2 : R / (‖(ρ : ℂ)‖ * (R - ‖(ρ : ℂ)‖)) ≤ 2 / (R - ‖(ρ : ℂ)‖) := by\n rw [div_mul_eq_mul_div, div_le_iff₀ (by linarith [hd2] : (0:ℝ) < R - ‖(ρ:ℂ)‖)]\n ...\n```\nAlternative cleaner: multiply both sides: R/(‖ρ‖·d) ≤ 2/d ⟺ R·d ≤ 2·(‖ρ‖·d) (mult by d > 0... and divide by d): R/(‖ρ‖·d) = (R/‖ρ‖)/d: then (R/‖ρ‖)/d ≤ 2/d ⟸ R/‖ρ‖ ≤ 2 (div_le_div_of_nonneg_right with d > 0). R/‖ρ‖ ≤ 2 ⟺ R ≤ 2‖ρ‖ ✓ (h5). So:\n```\nhave h2a : R / ‖(ρ : ℂ)‖ ≤ 2 := by\n rw [div_le_iff₀ hρpos]\n linarith [h5] -- R ≤ 2‖ρ‖: h5 : R ≤ 2·‖ρ‖ ✓... div_le_iff₀ (hb : 0 < b) : a / b ≤ c ↔ a ≤ c * b: R ≤ 2·‖ρ‖ ✓ h5\nhave h2 : R / (‖(ρ : ℂ)‖ * (R - ‖(ρ : ℂ)‖)) ≤ 2 / (R - ‖(ρ : ℂ)‖) := by\n rw [show R / (‖(ρ : ℂ)‖ * (R - ‖(ρ : ℂ)‖)) = (R / ‖(ρ : ℂ)‖) / (R - ‖(ρ : ℂ)‖) from by rw [div_div]]\n exact div_le_div_of_nonneg_right h2a (by linarith [hd2] : (0 : ℝ) < R - ‖(ρ : ℂ)‖)\n```\nHmm — div_le_div_of_nonneg_right : a ≤ b → 0 < c → a/c ≤ b/c? The name in current Mathlib: `div_le_div_of_nonneg_right (hab : a ≤ b) (hc : 0 < c) : a / c ≤ b / c`? I recall `div_le_div_of_nonneg_right` hmm — there's div_le_div_right (ordered field): `div_le_div_right (hc : 0 < c) : a / c ≤ b / c ↔ a ≤ b`. So: `(div_le_div_right (by linarith [hd2] : (0:ℝ) < R - ‖(ρ:ℂ)‖)).mpr h2a` ✓ robust.\ndiv_div : a / (b * c)?? div_div : a / b / c = a / (b * c) — so a/(b·c) = a/b/c via (div_div a b c).symm. My show: R/(‖ρ‖·d) = (R/‖ρ‖)/d: `rw [div_div]` on (R/‖ρ‖)/d gives R/(‖ρ‖·d) — so the show's proof is `by rw [div_div]` ✓.\n\nAlso check div_le_iff₀ exists (used in shell_low h3 — that compiled ✓ `div_le_iff₀ (by linarith [hR] : (0 : ℝ) < R - 3)` gave R ≤ 2·(R−3) ✓ worked).\n\nLet me fix h2.","signature":""},{"type":"text","text":"`div_le_div_iff₀`'s form mismatched. Rerouting h2 via `R/‖ρ‖ ≤ 2` + `div_le_div_right`."},{"type":"tool_use","id":"Edit_93","name":"Edit","input":{"replace_all":false,"file_path":"D:\\Python\\Quantum\\research\\lean\\NguyenRH1\\NguyenRH1\\LiCircleBound.lean","old_string":" have h2 : R / (‖(ρ : ℂ)‖ * (R - ‖(ρ : ℂ)‖)) ≤ 2 / (R - ‖(ρ : ℂ)‖) := by\n rw [div_le_div_iff₀ (mul_pos hρpos (by linarith [hd2] :\n (0 : ℝ) < ‖(ρ : ℂ)‖ * (R - ‖(ρ : ℂ)‖)))\n (by linarith [hd2] : (0 : ℝ) < R - ‖(ρ : ℂ)‖)]\n have h5 : R ≤ 2 * ‖(ρ : ℂ)‖ := by linarith [hρR2']\n have h6 : (0 : ℝ) ≤ R - ‖(ρ : ℂ)‖ := by linarith [hd2]\n nlinarith [h5, h6, hρpos]","new_string":" have h2 : R / (‖(ρ : ℂ)‖ * (R - ‖(ρ : ℂ)‖)) ≤ 2 / (R - ‖(ρ : ℂ)‖) := by\n have h5 : R ≤ 2 * ‖(ρ : ℂ)‖ := by linarith [hρR2']\n have h2a : R / ‖(ρ : ℂ)‖ ≤ 2 := by\n rw [div_le_iff₀ hρpos]\n linarith [h5]\n rw [show R / (‖(ρ : ℂ)‖ * (R - ‖(ρ : ℂ)‖))\n = (R / ‖(ρ : ℂ)‖) / (R - ‖(ρ : ℂ)‖) from by rw [div_div]]\n exact (div_le_div_right (by linarith [hd2] : (0 : ℝ) < R - ‖(ρ : ℂ)‖)).mpr h2a"},"extra_content":null}]