Nuprl Lemma : member-fpf-vals2

[A:Type]. ∀[eq:EqDecider(A)]. ∀[B:A ⟶ Type]. ∀[P:A ⟶ 𝔹]. ∀[f:x:A fp-> B[x]]. ∀[x:{a:A| ↑(P a)} ]. ∀[v:B[x]].
  {(↑x ∈ dom(f)) ∧ (v f(x) ∈ B[x])} supposing (<x, v> ∈ fpf-vals(eq;P;f))


Proof




Definitions occuring in Statement :  fpf-vals: fpf-vals(eq;P;f) fpf-ap: f(x) fpf-dom: x ∈ dom(f) fpf: a:A fp-> B[a] l_member: (x ∈ l) deq: EqDecider(T) assert: b bool: 𝔹 uimplies: supposing a uall: [x:A]. B[x] guard: {T} so_apply: x[s] and: P ∧ Q set: {x:A| B[x]}  apply: a function: x:A ⟶ B[x] pair: <a, b> product: x:A × B[x] universe: Type equal: t ∈ T
Definitions unfolded in proof :  uall: [x:A]. B[x] uimplies: supposing a member: t ∈ T all: x:A. B[x] iff: ⇐⇒ Q and: P ∧ Q implies:  Q prop: so_apply: x[s] so_lambda: λ2x.t[x] subtype_rel: A ⊆B guard: {T} top: Top

Latex:
\mforall{}[A:Type].  \mforall{}[eq:EqDecider(A)].  \mforall{}[B:A  {}\mrightarrow{}  Type].  \mforall{}[P:A  {}\mrightarrow{}  \mBbbB{}].  \mforall{}[f:x:A  fp->  B[x]].  \mforall{}[x:\{a:A|  \muparrow{}(P  a)\}  ].
\mforall{}[v:B[x]].
    \{(\muparrow{}x  \mmember{}  dom(f))  \mwedge{}  (v  =  f(x))\}  supposing  (<x,  v>  \mmember{}  fpf-vals(eq;P;f))



Date html generated: 2016_05_16-AM-11_18_27
Last ObjectModification: 2015_12_29-AM-09_21_51

Theory : event-ordering


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