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Answer a family under the condition it carries - #2206
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Rafael-SOWNet merged 1 commit intoOct 29, 2026
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sqrt(y) = x in x and y was declined. Solved for y it is x^2 provided x is 0 or -pi/2 < arg(x) <= pi/2, where sqrt takes its values, and the family (t_1, t_1^2 provided ...) was checked as though it claimed every t_1: sqrt(t_1^2) - t_1 is not zero, so the system was declined. A family that says where it holds is now checked there. The non-square solver takes the conditions out of the family's entries and decides forall t_1 in CC : condition implies residual = 0. Under that hypothesis a new reader takes the principal root of a whole power, (t^n)^(1/n), as t where the facts in scope put t at 0 or in the sector -pi/n < arg(t) <= pi/n, which is how the power's inversion writes it, or on t >= 0, as Complex.pow_cpow_nat_inv has it. A condition that Simplify introduces is not the family's, and is checked as before. sqrt(y) - x in x, y is [[t_1, t_1^2 provided t_1 = 0 or -1/2 * pi < arg(t_1) and arg(t_1) <= 1/2 * pi]], and y^(1/3) - x likewise with the cube root's sector. Part of #212 Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
Codecov Report❌ Patch coverage is Additional details and impacted files@@ Coverage Diff @@
## master #2206 +/- ##
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- Coverage 86.57% 86.57% -0.01%
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Files 301 301
Lines 66838 66878 +40
Branches 19410 19428 +18
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+ Hits 57864 57897 +33
Misses 4879 4879
- Partials 4095 4102 +7
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Part of #212: the family of
sqrt(y) - x, the first of the items left there.sqrt(y) = xinxandywas declined. Solved fory, it isx^2 provided x = 0 or -pi/2 < arg(x) <= pi/2, wheresqrttakes its values (#2183). The family(t_1, t_1^2 provided …)was then checked as though it claimed everyt_1, andsqrt(t_1^2) - t_1is not zero.forall t_1 in CC : condition implies residual = 0. A condition thatSimplifyintroduces is not the family's, and is checked as before.QuantifierFactstakes the principal root of a whole power,(t^n)^(1/n)forn >= 2, ast. That holds where the facts puttat0or in the sector-pi/n < arg(t) <= pi/n, which is how the inversion writes it, or ont >= 0, asComplex.pow_cpow_nat_invhas it. Outside a quantifiersqrt(t^2)is nott.sqrt(y) - xinx, ynull[[t_1, t_1 ^ 2 provided t_1 = 0 or -1/2 * pi < arg(t_1) and arg(t_1) <= 1/2 * pi]]y^(1/3) - xinx, ynull1/3 * piforall t in RR : t >= 0 implies sqrt(t^2) = tTrueforall t in CC : (t = 0 or -pi/2 < arg(t) <= pi/2) implies sqrt(t^2) = tTrueThe last item on #212 is the parameters' names in the result, an API question for the thread.
Tests:
RootByFactsTest(4 decided, 3 not claimed, one of them the square root's sector asked of a cube root, and nothing assumed outside a quantifier) andAFamilyThatSaysWhereItHoldsIsCheckedThere. The latter checks the family at points in the sector,i/2on its edge among them, and that the condition is false at a negativet_1. Full suite 17868/0, interactive trio, AOT smoke 64/0, all frameworks build.🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura