Nuprl Lemma : subtype-fpf-general
∀[A:Type]. ∀[P:A ⟶ ℙ]. ∀[B:A ⟶ 𝕌{j}]. (a:{a:A| P[a]} fp-> B[a] ⊆r a:A fp-> B[a])
Proof
Definitions occuring in Statement :
fpf: a:A fp-> B[a],
subtype_rel: A ⊆r B,
uall: ∀[x:A]. B[x],
prop: ℙ,
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,
subtype_rel: A ⊆r B,
fpf: a:A fp-> B[a],
so_apply: x[s],
uimplies: b supposing a,
so_lambda: λ2x.t[x],
all: ∀x:A. B[x],
implies: P ⇒ Q,
iff: P ⇐⇒ Q,
and: P ∧ Q,
rev_implies: P ⇐ Q,
prop: ℙ
Latex:
\mforall{}[A:Type]. \mforall{}[P:A {}\mrightarrow{} \mBbbP{}]. \mforall{}[B:A {}\mrightarrow{} \mBbbU{}\{j\}]. (a:\{a:A| P[a]\} fp-> B[a] \msubseteq{}r a:A fp-> B[a])
Date html generated:
2016_05_16-AM-11_02_43
Last ObjectModification:
2015_12_29-AM-09_14_16
Theory : event-ordering
Home
Index