First-order logic cannot count: games on bare sets
Lax945089.GamesOnSets · concepts/Lax945089/GamesOnSets.lean · lax-945089
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Theorem
Two finite sets with at least elements each, as structures over the empty vocabulary, are -round equivalent: the duplicator answers a new element by a new element and an old one by its match. Hence an order-free first-order definable property of bare sets is constant on the sets beyond some size: first-order logic counts up to its quantifier depth and no further.
Concept map
Evidence
This concept declares 2 statements. Each proof establishes one of them relative to its assumptions.
1 efEquiv_bare proven
2 exists_card_bound_of_foDefinableFree proven
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Builds on
Lax134656.PartialFixedPointLax485149.ClassLLax485149.ClassNLLax485149.DeterministicTransitiveClosureLax485149.FirstOrderDefinabilityLax485149.ProblemsLax485149.TransitiveClosureLax535992.ClassPTIMELax535992.InflationaryFixedPointLax895169.ArithmeticLogicLax904597.ClassesLax904597.InterpretationsLax904597.ProblemsLax904597.RelativizedLax904597.SecondOrderLax945089.EhrenfeuchtGamesLax945089.EvenLax945089.OrderFreeFirstOrderLax945089.ParityLax945089.PebbleGamesLax945089.TransitiveClosureReductions
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