Nuprl Lemma : test5
∀[A:Type]. ∀[B:A ⟶ Type]. ∀[f:a:A ⟶ B[a]]. ∀[x:A]. (f x ∈ B[x])
Proof
Definitions occuring in Statement :
uall: ∀[x:A]. B[x],
so_apply: x[s],
member: t ∈ T,
apply: f a,
function: x:A ⟶ B[x],
universe: Type
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
so_apply: x[s]
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
isect_memberFormation_alt,
cut,
applyEquality,
hypothesisEquality,
sqequalHypSubstitution,
hypothesis,
universeIsType,
functionIsType,
universeEquality
Latex:
\mforall{}[A:Type]. \mforall{}[B:A {}\mrightarrow{} Type]. \mforall{}[f:a:A {}\mrightarrow{} B[a]]. \mforall{}[x:A]. (f x \mmember{} B[x])
Date html generated:
2019_10_15-AM-11_36_32
Last ObjectModification:
2018_10_16-PM-00_29_36
Theory : general
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