`bool_hasIdentityStep`

proved

show as: view math explainer →

module: IndisputableMonolith.Foundation.DiscreteLogicRealization
domain: Foundation
line: 67 · github
papers citing: none yet

open explainer

Generate a durable explainer page for this declaration.

open lean source

IndisputableMonolith.Foundation.DiscreteLogicRealization on GitHub at line 67.

browse module

All declarations in this module, on Recognition.

explainer page

Tracked in the explainer inventory; generation is lazy so crawlers do not trigger LLM jobs.

open explainer

depends on

formal source

  64    simp [boolCost]
  65
  66/-- The discrete realization has a non-trivial identity-step shadow. -/
  67theorem bool_hasIdentityStep : boolRealization.hasIdentityStep :=
  68  LogicRealization.hasIdentityStep_of_nontrivial boolRealization
  69
  70/-- Boolean realization has the same forced arithmetic as every realization. -/
  71noncomputable def bool_arithmetic_invariant (R : LogicRealization.{0, 0}) :
  72    (UniversalForcing.arithmeticOf boolRealization).peano.carrier
  73      ≃ (UniversalForcing.arithmeticOf R).peano.carrier :=
  74  ArithmeticOf.equivOfInitial
  75    (UniversalForcing.arithmeticOf boolRealization) (UniversalForcing.arithmeticOf R)
  76
  77/-- The Boolean realization's forced arithmetic has the Peano surface. -/
  78theorem bool_peano_surface :
  79    ArithmeticOf.PeanoSurface (UniversalForcing.arithmeticOf boolRealization) :=
  80  UniversalForcing.peano_surface boolRealization
  81
  82end DiscreteLogicRealization
  83end Foundation
  84end IndisputableMonolith

papers checked against this theorem

No papers in the current corpus map onto this theorem yet.