Nuprl Lemma : respects-equality-set-trivial2

[T:Type]. ∀[P:T ⟶ ℙ].  respects-equality({x:T| P[x]} ;T)


Proof




Definitions occuring in Statement :  respects-equality: respects-equality(S;T) uall: [x:A]. B[x] prop: so_apply: x[s] set: {x:A| B[x]}  function: x:A ⟶ B[x] universe: Type
Definitions unfolded in proof :  uall: [x:A]. B[x] member: t ∈ T respects-equality: respects-equality(S;T) all: x:A. B[x] implies:  Q prop: so_apply: x[s] squash: T
Lemmas referenced :  istype-base
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity Error :isect_memberFormation_alt,  introduction cut Error :lambdaFormation_alt,  hypothesis sqequalRule Error :equalityIstype,  because_Cache hypothesisEquality sqequalBase equalitySymmetry Error :setIsType,  Error :universeIsType,  applyEquality Error :inhabitedIsType,  extract_by_obid sqequalHypSubstitution Error :lambdaEquality_alt,  dependent_functionElimination thin axiomEquality Error :functionIsTypeImplies,  Error :functionIsType,  universeEquality Error :isect_memberEquality_alt,  isectElimination Error :isectIsTypeImplies,  applyLambdaEquality setElimination rename imageMemberEquality baseClosed imageElimination

Latex:
\mforall{}[T:Type].  \mforall{}[P:T  {}\mrightarrow{}  \mBbbP{}].    respects-equality(\{x:T|  P[x]\}  ;T)



Date html generated: 2019_06_20-AM-11_13_43
Last ObjectModification: 2018_12_02-PM-11_41_35

Theory : core_2


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