Nuprl Lemma : b-union-base-equality
∀A:Type. ∀a,b:Base.  ((a = b ∈ (Base ⋃ A)) 
⇒ (b ∈ A) 
⇒ (a = b ∈ A))
Proof
Definitions occuring in Statement : 
b-union: A ⋃ B
, 
all: ∀x:A. B[x]
, 
implies: P 
⇒ Q
, 
member: t ∈ T
, 
base: Base
, 
universe: Type
, 
equal: s = t ∈ T
Definitions unfolded in proof : 
all: ∀x:A. B[x]
, 
uall: ∀[x:A]. B[x]
, 
member: t ∈ T
, 
and: P ∧ Q
, 
implies: P 
⇒ Q
, 
guard: {T}
Lemmas referenced : 
strong-subtype-union-base, 
strong-subtype-implies, 
b-union_wf, 
base_wf, 
istype-universe, 
istype-base
Rules used in proof : 
cut, 
introduction, 
extract_by_obid, 
sqequalSubstitution, 
sqequalTransitivity, 
computationStep, 
sqequalReflexivity, 
Error :lambdaFormation_alt, 
hypothesis, 
sqequalHypSubstitution, 
isectElimination, 
thin, 
hypothesisEquality, 
productElimination, 
independent_functionElimination, 
instantiate, 
universeEquality, 
dependent_functionElimination, 
equalityTransitivity, 
equalitySymmetry, 
sqequalRule, 
Error :equalityIstype, 
Error :universeIsType, 
sqequalBase, 
because_Cache
Latex:
\mforall{}A:Type.  \mforall{}a,b:Base.    ((a  =  b)  {}\mRightarrow{}  (b  \mmember{}  A)  {}\mRightarrow{}  (a  =  b))
Date html generated:
2019_06_20-PM-00_28_06
Last ObjectModification:
2018_12_07-AM-11_43_12
Theory : subtype_1
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