[{"type":"thinking","thinking":"The `by have h1 : N₀ ≤ y := hy.1; linarith` inside the term-mode lambda: the `by` block's goal is `a < y` — linarith needs hN : a < N₀ and h1 : N₀ ≤ y — but linarith says failed to find contradiction with goal False?? The context shows `⊢ False` — that means the by-block's goal is being shown as False? No wait — the error context shows `h1 : N₀ ≤ y ⊢ False` — the goal displayed is False, meaning linarith got called on a goal it already... Hmm, \"linarith failed to find a contradiction\" with ⊢ False means linarith was asked to prove False from the hypotheses (that's how linarith works when the goal isn't an inequality — it fails). But the goal should be `a < y`. The `by have h1...; linarith` term inside `(hdiff y (...))` — the expected type of the `...` is `y ∈ Set.Ioi a` which is `a < y` — should be an inequality. But the context shows ⊢ False?! \n\nOh wait — I see: `hdiff y (by have h1 : N₀ ≤ y := hy.1; linarith)` — the `by` block might be parsed as applying to the whole expression... no. Actually the issue: `Set.Ioi a` membership `y ∈ Set.Ioi a` — maybe it elaborates with a DIFFERENT y? No...\n\nHmm, actually look: the error is at 2336:45 and shows goal ⊢ False with hy : y ∈ Set.Icc N₀ x, h1 : N₀ ≤ y in context. linarith failed to find contradiction for False. That means the by-block goal IS False — no wait, if the goal were a < y, linarith would prove it or fail with \"failed to find contradiction\" showing goal a < y... Actually linarith ALWAYS works by contradiction: it negates the goal and tries to derive False. When it FAILS, the error shows the negated context with ⊢ False. The context shown includes hy, h1, and the NEGATED goal should appear as a hypothesis (a < y pushed as ¬(a < y) i.e. y ≤ a)... but the context shows NO hypothesis about a and y! hN : a < N₀ is there. h1 : N₀ ≤ y is there. So from a < N₀ ≤ y, linarith should derive a < y — trivial. Unless... the context shown is the FULL context and linarith still failed?\n\nWait, look more carefully at the first error — it's cut off at the top (I only see from `f : ℝ → ℝ`). The negated goal would be `a✝ : y ≤ a` or similar — I don't see it. Hmm, but the context shows hN, hx, h0, h0', hlt, y, hy, h1 — and ⊢ False. With hN : a < N₀ and h1 : N₀ ≤ y: linarith derives a < y; with negated goal y ≤ a it derives False — should WORK. Unless the negated goal isn't in the displayed context because... the display is truncated? The first error's context start is cut (the tail -30 cut it). Possibly the missing piece: hy.1 for y ∈ Set.Icc N₀ x gives N₀ ≤ y ✓ that's h1. And goal... \n\nHmm wait — maybe the issue: `Set.Icc N₀ x` membership hy.1 : N₀ ≤ y ✓. The by-block goal: `y ∈ Set.Ioi a`. `y ∈ Set.Ioi a` is NOT definitionally `a < y` for linarith? It IS: Set.mem_Ioi : y ∈ Ioi a ↔ a < y, and ∈ Ioi is reducible to a < y. linarith should see through it (mem Ioi is defeq to a < y). But the displayed ⊢ False suggests linarith's preprocessing turned the goal... \n\nOH WAIT. I see it now! The error is that in the FIRST have (hfc), the `by` block inside the term `(hdiff y (by have h1 : N₀ ≤ y := hy.1; linarith))` — hy.1 where hy : y ∈ Set.Icc N₀ x — `hy.1` works on membership of Icc? y ∈ Icc N₀ x is N₀ ≤ y ∧ y ≤ x (structure And) — hy.1 : N₀ ≤ y ✓. Then linarith with goal a < y... \n\nHmm, let me reconsider: maybe the problem is that `linarith` can't see `hN`? hN : a < N₀ IS in context. a < y from a < N₀ ≤ y — one step. linarith definitely closes this. Unless the goal ISN'T a < y.\n\nLook at error 2 (2336:45): context has hfc already (so this is hfd's block), hy : y ∈ Set.Ioo N₀ x, h1 : N₀ < y, ⊢ False. Again missing the negated goal in display... and again a < y from a < N₀ < y.\n\nHold on — maybe the negated-goal hypothesis IS there but the tail cut it. Both errors at column 45: `by have h1 : N₀ ≤ y := hy.1; linarith` — column 45 is around `linarith`. The negated goal would appear as a hypothesis `a✝ : y ≤ a` — not visible in either context dump! So the goal being proved is genuinely `False`?? How?\n\nOHHH. I bet the issue: `Set.Ioi a` — hdiff expects `y ∈ Set.Ioi a`. The by-block: `by have h1 : N₀ ≤ y := hy.1; linarith`. If linarith is called with goal `y ∈ Set.Ioi a`: linarith preprocesses — it should unfold membership... Actually linarith does handle ∈ Ioi goals via `simp` preprocessing? Not reliably! linarith's preprocessing includes `simp only [Set.mem_Ioi,...]`? I don't think membership goals are handled. But then the goal would be `y ∈ Set.Ioi a` and linarith would fail with \"linarith failed to find a contradiction\" after negating — negation of a Prop membership isn't an inequality linarith understands... it might just fail to even negate properly, showing ⊢ False.\n\nHmm, but the earlier proof (hdiffM application in lambda_pos_finite_height) used `Ioi_mem_nhds hy` — different. And in marginTilde_convex, `hmem m hm` used Ioi_mem_nhds. For membership goals the file used explicit Set.mem_Ioi.mpr. So YES: the goal y ∈ Set.Ioi a needs Set.mem_Ioi.mpr wrapping: linarith proves a < y, then Set.mem_Ioi.mpr. Fix:\n\n```\n(hdiff y (Set.mem_Ioi.mpr (by have h1 : N₀ ≤ y := hy.1; linarith))).differentiableWithinAt\n```\nWait but if goal of by-block is `y ∈ Set.Ioi a` — defeq to `a < y` — does `by ... linarith` see it as a < y? linarith calls `getGoalType` and checks if it's an inequality after `whnfR`? Evidently not, given the failure. So wrap with Set.mem_Ioi.mpr and give linarith the a < y goal directly.\n\nFix both:\n```\n    have hfc : ContinuousOn f (Set.Icc N₀ x) :=\n      DifferentiableOn.continuousOn (s := Set.Icc N₀ x) (f := f) fun y hy =>\n        (hdiff y (Set.mem_Ioi.mpr (by\n          have h1 : N₀ ≤ y := hy.1\n          linarith))).differentiableWithinAt\n    have hfd : DifferentiableOn ℝ f (Set.Ioo N₀ x) := fun y hy =>\n      (hdiff y (Set.mem_Ioi.mpr (by\n        have h1 : N₀ < y := hy.1\n        linarith))).differentiableWithinAt\n```\n\nAlso check whether there were more errors ABOVE these (the tail -30 cut the beginning). Let me recompile and get full error list.","signature":""},{"type":"text","text":"The `∈ Set.Ioi` membership goals need `Set.mem_Ioi.mpr` wrapping (linarith can't see through membership). Fixing."},{"type":"tool_use","id":"Edit_67","name":"Edit","input":{"file_path":"D:\\Python\\Quantum\\research\\lean\\NguyenRH1\\NguyenRH1\\LiSkeleton.lean","old_string":"  · have hfc : ContinuousOn f (Set.Icc N₀ x) :=\n      DifferentiableOn.continuousOn (s := Set.Icc N₀ x) (f := f) fun y hy =>\n        (hdiff y (by have h1 : N₀ ≤ y := hy.1; linarith)).differentiableWithinAt\n    have hfd : DifferentiableOn ℝ f (Set.Ioo N₀ x) := fun y hy =>\n      (hdiff y (by have h1 : N₀ < y := hy.1; linarith)).differentiableWithinAt","new_string":"  · have hfc : ContinuousOn f (Set.Icc N₀ x) :=\n      DifferentiableOn.continuousOn (s := Set.Icc N₀ x) (f := f) fun y hy =>\n        (hdiff y (Set.mem_Ioi.mpr (by\n          have h1 : N₀ ≤ y := hy.1\n          linarith))).differentiableWithinAt\n    have hfd : DifferentiableOn ℝ f (Set.Ioo N₀ x) := fun y hy =>\n      (hdiff y (Set.mem_Ioi.mpr (by\n        have h1 : N₀ < y := hy.1\n        linarith))).differentiableWithinAt"},"extra_content":null}]