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Nested quantifiers are decided through listed witnesses and witness terms - #1564
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…erms exists z in S : P, where the solver lists the z at which P holds and the quantifiers around it make each a whole number, is the statement that one of them is in S, a comparison in ZZ* and ZZ+: Sullivan and Mackey's Prob 4.11.6 is True so. Two rays with symbolic ends meet, so that x < y and y < x is solved to nothing. And exists x : forall y : P is False where one of a few fixed terms in x fails P at every x, forall x : exists y : P True where one holds P at every x: Prob 4.11.5 fails at y = x^2 + 1. Part of #1409. Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
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Two of Sullivan and Mackey's nested-quantifier problems are decided, through
BySolvingas #1559 left it, and one more route.Prob 4.11.6:
forall x in ZZ : forall y in ZZ : exists z in ZZ* : x - y = z or y - x = zisTrue.x - yandy - x. Where the quantifiers around the statement make each witness a whole number,exists z in S : Pbecomes the statement that one of them is inS.QuantifierFacts.MembershipAsComparisonreads each membership:s in ZZ*iss >= 0ands in ZZ+iss >= 1. A quantifier over the whole numbers, and a membership in them, are now among the nodes whose body is decided with the facts in scope.ythen asks forx - y < 0 and y - x < 0, which needs two rays with symbolic ends to meet.(a; +oo) /\ (-oo; b)is(a; b), and(x; +oo) /\ (-oo; x)is empty. Both were left as written.Prob 4.11.5:
exists x in RR : forall y in RR : x^2 - y^2 >= 0isFalse.exists x : forall y : Pis false where one of a few fixed terms inxfailsPat everyx, andforall x : exists y : Pis true where one holdsPat everyx.y = x^2 + 1fails everywhere.LeftAsWrittenWhereNothingDecidesItstill are.Also: a comparison of two polynomials that differ by a number is decided at every member.
x^2 + 1 > x^2is1 > 0, which is what the witness termx^2 + 1asks.exists y in RR : forall x in RR : y = x^3moves out ofLeftAsWrittenWhereNothingDecidesIt: it isFalsenow, since whateveryis, the cube ofx = y^2 + 1is noty.Suite: 371 tests pass on net10.0 in the quantifier, interval, declined-statement, induction, surjectivity and inequality classes, and every target framework builds. The whole suite ran on this branch merged with the other open #1409 branches: 13,422 passed, 0 failed. Gate on that merge: passed, with allocation as the baseline says on all 19 gated benchmarks.
BREAKING-CHANGES.mdhas the entry, beside the one for threshold sets.Part of #1409.
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