← Session 35 — ii′-6-R: Lagarias 6.1 right-tail re-proof (E-032/D-073)
thinking (4590 chars)
Three clear issues:
1. **Line 1855 (marginTilde_eq_expanded's ring)**: the unsolved goal shows the expanded LHS with the constant `-320457/250` and `√n·(−996/5)` — the ring's normalization of the two sides differs. The constant −320457/250 = −1281.828 and −996/5 = −199.2 — so the LHS has mIdb's OLD coefficient (−199.2)?? — no wait, the displayed LHS is the NORMALIZED marginTilde form — and the constant −320457/250 is the mK's normalized constant — vs the expected marginExpanded's mK (−40·1.3352 − 2(e⁴+1)(7(3·2.9²+48)) − 16.76 − 1 + 1). The ring normalization shows mismatches — the blocks' arithmetic vs marginTilde's actual form. The issue: DogLegRight(n) = ((√n+1)/2)(e^{5/2}+1)/2 + 8.38 + IIMaj/4 — the 2·DogLegRight in marginTilde/Bn = (√n+1)(e^{5/2}+1) + 16.76 + IIMaj/2·... — wait: 2·DogLegRight(n) = 2·[((√n+1)/2)(e^{5/2}+1)/2 + 8.38 + IIMaj/4] = (√n+1)(e^{5/2}+1)/2·2/2 + 16.76 + 2·IIMaj/4 = (√n+1)(e^{5/2}+1)/2 + 16.76 + IIMaj/2·... hmm: 2·((√n+1)/2)·(e^{5/2}+1)/2 = (√n+1)·(e^{5/2}+1)/2·... — 2·((√n+1)/2)·((e^{5/2}+1)/2) = (√n+1)·(e^{5/2}+1)/2·2/2 = (√n+1)(e^{5/2}+1)/2·1 = (√n+1)(e^{5/2}+1)/2·1?? — let me just compute: 2·DogLegRight = 2·[((√n+1)/2)·(e^{5/2}+1)/2 + 8.38 + IIMaj/4] = 2·((√n+1)/2)·(e^{5/2}+1)/2 + 2·8.38 + 2·(IIMaj/4) = ((√n+1)·(e^{5/2}+1)/2)·... — 2·((√n+1)/2)·((e^{5/2}+1)/2) = (√n+1)·(e^{5/2}+1)?? NO: 2·((√n+1)/2)·((e^{5/2}+1)/2) = (√n+1)·((e^{5/2}+1)/2) = (√n+1)(e^{5/2}+1)/2·... — 2·((√n+1)/2) = (√n+1): then ×(e^{5/2}+1)/2: (√n+1)(e^{5/2}+1)/2·... — 2·((√n+1)/2)·(e^{5/2}+1)/2 = (√n+1)·(e^{5/2}+1)/2·2/2 — ugh, let me just: 2·[(√n+1)/2·(e^{5/2}+1)/2] = [(√n+1)·(e^{5/2}+1)]/2·... — (√n+1)/2 · (e^{5/2}+1)/2 = (√n+1)(e^{5/2}+1)/4. Times 2: (√n+1)(e^{5/2}+1)/2. So 2·DogLegRight = (√n+1)(e^{5/2}+1)/2 + 16.76 + IIMaj/2. Hmm wait: (√n+1)(e^{5/2}+1)/4·2 = (√n+1)(e^{5/2}+1)/2 ✓. So the √n-linear dog-leg coefficient in Bn is (e^{5/2}+1)/2 = 6.591 ✓ matches mIdb's added (e^{5/2}+1)/2 ✓.
So the √n-linear coefficient: mIdb's −[(8/3·2.7+192) + (e^{5/2}+1)/2] = −[199.2 + 6.591] = −205.791 — but the normalized goal shows `√n·(−996/5)` = −199.2 — that's the OLD coefficient!! So marginTilde_eq_expanded's ring compared the OLD mIdb-normalized form?? — no — the displayed goal is the result of ring normalizing BOTH sides — the LHS (marginTilde's actual) shows √n·(−996/5) = −199.2 — meaning marginTilde's actual √n coefficient is −199.2?! — but the new marginTilde/Bn has the dog-leg's −6.591 too... — wait, the displayed LHS has √n·(−996/5) = −199.2 AND the goal is UNSOLVED — the LHS lacks the −6.591 term. Where's the dog-leg √n-linear part in marginTilde's actual form? — the 2·DogLegRight in marginTilde: −2·DogLegRight(n) = −[(√n+1)(e^{5/2}+1)/2 + 16.76 + IIMaj/2] — the √n part: −(√n+1)(e^{5/2}+1)/2·... = −(e^{5/2}+1)/2·(√n+1) = −6.591·(√n+1) — that's −6.591·√n − 6.591. So marginTilde's actual √n coefficient should be −199.2 − 6.591 = −205.791 ✓ — but the normalized LHS shows −199.2?! — meaning the 2·DogLegRight's √n part ISN'T being unfolded in marginTilde_eq_expanded's simp only list. The eq_expanded's simp only list includes the block names (mFblock, mTL1b, ..., mIIM, mK, Hhat, C0hat, Z2hat) but NOT DogLegRight!! — so marginTilde's `2 * DogLegRight n` is NOT unfolded by the simp only, and ring compares with it opaque on one side. Fix: add DogLegRight to the simp only list in marginTilde_eq_expanded.
2. **Line 1938 (continuousOn_marginExpanded)**: Type mismatch on ContinuousOn.add — the mIdb block's form changed to `(-(((8/3)·2.7+192) + (e^{5/2}+1)/2))·(√m+1)` and the proof's `(hT.const_mul _)` — hT : ContinuousOn (fun m => √m+1) — .const_mul _ — the `_` should infer the new constant — the mismatch might be because the new coefficient is a SUM (199.2 + (e^{5/2}+1)/2) and the const_mul's inferred constant form differs. Let me look at the mIdb handling in continuousOn_marginExpanded (line 1930: `(hT.const_mul _))` — should work with any constant... — "Type mismatch ContinuousOn.add" — probably a cascade of error 1's elaboration.
3. **Lines 2279/2311 (vId/vIIM's rw)**:
- vId's rw: `show (fun m => -(8/3·2.7 + 192)·(√m+1)) = (fun m => -(8/3·2.7+192)·id(√m+1)) from rfl` — but mIdb is now `(-(((8/3)·2.7+192) + (e^{5/2}+1)/2))·(√m+1)` — the show's literal has the OLD coefficient −(8/3·2.7+192) = −199.2 — mismatch with the new coefficient. Fix the show's literal to the new coefficient.
- vIIM's rw: `deriv2_const_mul (-1) ...` — mIIM is now (−1.5)·IIMaj — the rw expects (−1) — change to (−1.5) and the value's literal.
Let me look at vId and vIIM's exact current text to fix. Also marginTilde_eq_expanded's simp list.
Three fixes: `DogLegRight` into the eq_expanded simp list, and the vId/vIIM literal updates. Let me look at the exact texts.
tool: Read ok
input
{
"file_path": "D:\\Python\\Quantum\\research\\lean\\NguyenRH1\\NguyenRH1\\LiSkeleton.lean",
"offset": 1842,
"limit": 22
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result 2026-08-26 11:59:35Z → 2026-08-26 11:59:35Z
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"content": " ((fun t => Real.log (Real.log (t + 2))) (Real.sqrt m + 1))\nprivate def mIIM : \u211d \u2192 \u211d := fun m => (-1.5) * IIMaj m\nprivate def mK : \u211d := -40*1.3352 - 2*(Real.exp 4 + 1)*(7*(3*2.9^2 + 48)) - 16.76 - 1 + 1\n\n/-- The margin as an explicit left-nested sum of pieces. -/\nprivate def marginExpanded (F : \u211d \u2192 \u211d) : \u211d \u2192 \u211d := fun n =>\n (((((((((((fun m => (1/2)*m*Real.log m + (lagariasC1 - Real.eulerMascheroniConstant/2)*m - 4) n +\n mFblock F n) + mTL1b n) + mTL2b n) + mIdb n) + mL1sqb n) + mL1L2b n) +\n mL2sqb n) + mL1b n) + mL2b n) + mIIM n) + mK\n\n/-- [PROVED session 011] marginTilde = marginExpanded (the D-056/D-057\nconstant bookkeeping, checked by `ring`). -/\ntheorem marginTilde_eq_expanded (F : \u211d \u2192 \u211d) :\n marginTilde F = marginExpanded F := by\n funext n\n simp only [marginTilde, marginExpanded, mFblock, mTL1b, mTL2b, mIdb, mL1sqb,\n mL1L2b, mL2sqb, mL1b, mL2b, mIIM, mK, Hhat, C0hat, Z2hat]\n ring\n\n/-- [PROVED session 011] The pieces are C\u00b2 on (20, \u221e). -/\ntheorem contDiffOn_mS (c : \u211d) :\n ContDiffOn \u211d 2 (fun m => (1/2)*m*Real.log m + c*m - 4) (Set.Ioi 20) :=",
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raw content_json (database row)
[{"type":"thinking","thinking":"Three clear issues:\n\n1. **Line 1855 (marginTilde_eq_expanded's ring)**: the unsolved goal shows the expanded LHS with the constant `-320457/250` and `√n·(−996/5)` — the ring's normalization of the two sides differs. The constant −320457/250 = −1281.828 and −996/5 = −199.2 — so the LHS has mIdb's OLD coefficient (−199.2)?? — no wait, the displayed LHS is the NORMALIZED marginTilde form — and the constant −320457/250 is the mK's normalized constant — vs the expected marginExpanded's mK (−40·1.3352 − 2(e⁴+1)(7(3·2.9²+48)) − 16.76 − 1 + 1). The ring normalization shows mismatches — the blocks' arithmetic vs marginTilde's actual form. The issue: DogLegRight(n) = ((√n+1)/2)(e^{5/2}+1)/2 + 8.38 + IIMaj/4 — the 2·DogLegRight in marginTilde/Bn = (√n+1)(e^{5/2}+1) + 16.76 + IIMaj/2·... — wait: 2·DogLegRight(n) = 2·[((√n+1)/2)(e^{5/2}+1)/2 + 8.38 + IIMaj/4] = (√n+1)(e^{5/2}+1)/2·2/2 + 16.76 + 2·IIMaj/4 = (√n+1)(e^{5/2}+1)/2 + 16.76 + IIMaj/2·... hmm: 2·((√n+1)/2)·(e^{5/2}+1)/2 = (√n+1)·(e^{5/2}+1)/2·... — 2·((√n+1)/2)·((e^{5/2}+1)/2) = (√n+1)·(e^{5/2}+1)/2·2/2 = (√n+1)(e^{5/2}+1)/2·1 = (√n+1)(e^{5/2}+1)/2·1?? — let me just compute: 2·DogLegRight = 2·[((√n+1)/2)·(e^{5/2}+1)/2 + 8.38 + IIMaj/4] = 2·((√n+1)/2)·(e^{5/2}+1)/2 + 2·8.38 + 2·(IIMaj/4) = ((√n+1)·(e^{5/2}+1)/2)·... — 2·((√n+1)/2)·((e^{5/2}+1)/2) = (√n+1)·(e^{5/2}+1)?? NO: 2·((√n+1)/2)·((e^{5/2}+1)/2) = (√n+1)·((e^{5/2}+1)/2) = (√n+1)(e^{5/2}+1)/2·... — 2·((√n+1)/2) = (√n+1): then ×(e^{5/2}+1)/2: (√n+1)(e^{5/2}+1)/2·... — 2·((√n+1)/2)·(e^{5/2}+1)/2 = (√n+1)·(e^{5/2}+1)/2·2/2 — ugh, let me just: 2·[(√n+1)/2·(e^{5/2}+1)/2] = [(√n+1)·(e^{5/2}+1)]/2·... — (√n+1)/2 · (e^{5/2}+1)/2 = (√n+1)(e^{5/2}+1)/4. Times 2: (√n+1)(e^{5/2}+1)/2. So 2·DogLegRight = (√n+1)(e^{5/2}+1)/2 + 16.76 + IIMaj/2. Hmm wait: (√n+1)(e^{5/2}+1)/4·2 = (√n+1)(e^{5/2}+1)/2 ✓. So the √n-linear dog-leg coefficient in Bn is (e^{5/2}+1)/2 = 6.591 ✓ matches mIdb's added (e^{5/2}+1)/2 ✓.\n\nSo the √n-linear coefficient: mIdb's −[(8/3·2.7+192) + (e^{5/2}+1)/2] = −[199.2 + 6.591] = −205.791 — but the normalized goal shows `√n·(−996/5)` = −199.2 — that's the OLD coefficient!! So marginTilde_eq_expanded's ring compared the OLD mIdb-normalized form?? — no — the displayed goal is the result of ring normalizing BOTH sides — the LHS (marginTilde's actual) shows √n·(−996/5) = −199.2 — meaning marginTilde's actual √n coefficient is −199.2?! — but the new marginTilde/Bn has the dog-leg's −6.591 too... — wait, the displayed LHS has √n·(−996/5) = −199.2 AND the goal is UNSOLVED — the LHS lacks the −6.591 term. Where's the dog-leg √n-linear part in marginTilde's actual form? — the 2·DogLegRight in marginTilde: −2·DogLegRight(n) = −[(√n+1)(e^{5/2}+1)/2 + 16.76 + IIMaj/2] — the √n part: −(√n+1)(e^{5/2}+1)/2·... = −(e^{5/2}+1)/2·(√n+1) = −6.591·(√n+1) — that's −6.591·√n − 6.591. So marginTilde's actual √n coefficient should be −199.2 − 6.591 = −205.791 ✓ — but the normalized LHS shows −199.2?! — meaning the 2·DogLegRight's √n part ISN'T being unfolded in marginTilde_eq_expanded's simp only list. The eq_expanded's simp only list includes the block names (mFblock, mTL1b, ..., mIIM, mK, Hhat, C0hat, Z2hat) but NOT DogLegRight!! — so marginTilde's `2 * DogLegRight n` is NOT unfolded by the simp only, and ring compares with it opaque on one side. Fix: add DogLegRight to the simp only list in marginTilde_eq_expanded.\n\n2. **Line 1938 (continuousOn_marginExpanded)**: Type mismatch on ContinuousOn.add — the mIdb block's form changed to `(-(((8/3)·2.7+192) + (e^{5/2}+1)/2))·(√m+1)` and the proof's `(hT.const_mul _)` — hT : ContinuousOn (fun m => √m+1) — .const_mul _ — the `_` should infer the new constant — the mismatch might be because the new coefficient is a SUM (199.2 + (e^{5/2}+1)/2) and the const_mul's inferred constant form differs. Let me look at the mIdb handling in continuousOn_marginExpanded (line 1930: `(hT.const_mul _))` — should work with any constant... — \"Type mismatch ContinuousOn.add\" — probably a cascade of error 1's elaboration.\n\n3. **Lines 2279/2311 (vId/vIIM's rw)**: \n- vId's rw: `show (fun m => -(8/3·2.7 + 192)·(√m+1)) = (fun m => -(8/3·2.7+192)·id(√m+1)) from rfl` — but mIdb is now `(-(((8/3)·2.7+192) + (e^{5/2}+1)/2))·(√m+1)` — the show's literal has the OLD coefficient −(8/3·2.7+192) = −199.2 — mismatch with the new coefficient. Fix the show's literal to the new coefficient.\n- vIIM's rw: `deriv2_const_mul (-1) ...` — mIIM is now (−1.5)·IIMaj — the rw expects (−1) — change to (−1.5) and the value's literal.\n\nLet me look at vId and vIIM's exact current text to fix. Also marginTilde_eq_expanded's simp list.","signature":""},{"type":"text","text":"Three fixes: `DogLegRight` into the eq_expanded simp list, and the vId/vIIM literal updates. Let me look at the exact texts."},{"type":"tool_use","id":"Read_605","name":"Read","input":{"file_path":"D:\\Python\\Quantum\\research\\lean\\NguyenRH1\\NguyenRH1\\LiSkeleton.lean","offset":1842,"limit":22},"extra_content":null}]