[Merged by Bors] - chore(Analysis/SpecialFunctions/Integrals): simplify proof of intervalIntegrable_cpow #19877
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PR summary 153dff3bd5Import changes for modified filesNo significant changes to the import graph Import changes for all files
Declarations diffNo declarations were harmed in the making of this PR! 🐙 You can run this locally as follows## summary with just the declaration names:
./scripts/declarations_diff.sh <optional_commit>
## more verbose report:
./scripts/declarations_diff.sh long <optional_commit>The doc-module for No changes to technical debt.You can run this locally as
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dwrensha
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Dec 10, 2024
| rw [← Ioo_union_Icc_eq_Ioc hc (le_refl 0), ← Icc_def] | ||
| simp_rw [← le_antisymm_iff, setOf_eq_eq_singleton'] | ||
| rw [this, integrableOn_union, and_comm]; constructor | ||
| rw [← Ioo_union_right hc, integrableOn_union, and_comm]; constructor |
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Here, hc has type c < 0, and Ioo_union_right applied to it has type Set.Ioo c 0 ∪ {0} = Set.Ioc c 0.
(Just noting this to show that the change I'm proposing does not introduce any defeq abuse.)
grunweg
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🚀 Pull request has been placed on the maintainer queue by grunweg. |
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bors r+ |
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…lIntegrable_cpow (#19877) * `have : Ioc c 0 = Ioo c 0 ∪ {(0 : ℝ)} ` is directly proved by `(Ioo_union_right hc).symm` * That shorter proof can then be inlined at its use in the `simp` in the next line. This simplification was found by [`tryAtEachStep`](https://github.com/dwrensha/tryAtEachStep).
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Pull request successfully merged into master. Build succeeded: |
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have : Ioc c 0 = Ioo c 0 ∪ {(0 : ℝ)}is directly proved by(Ioo_union_right hc).symmsimpin the next line.This simplification was found by
tryAtEachStep.