Nuprl Lemma : fpf-sub_functionality

[A,A':Type].
  ∀[B:A ─→ Type]. ∀[C:A' ─→ Type]. ∀[eq:EqDecider(A)]. ∀[eq':EqDecider(A')]. ∀[f,g:a:A fp-> B[a]].
    {f ⊆ supposing f ⊆ g} supposing ∀a:A. (B[a] ⊆C[a]) 
  supposing strong-subtype(A;A')


Proof




Definitions occuring in Statement :  fpf-sub: f ⊆ g fpf: a:A fp-> B[a] deq: EqDecider(T) strong-subtype: strong-subtype(A;B) uimplies: supposing a subtype_rel: A ⊆B uall: [x:A]. B[x] guard: {T} so_apply: x[s] all: x:A. B[x] function: x:A ─→ B[x] universe: Type
Lemmas :  fpf-sub-functionality
\mforall{}[A,A':Type].
    \mforall{}[B:A  {}\mrightarrow{}  Type].  \mforall{}[C:A'  {}\mrightarrow{}  Type].  \mforall{}[eq:EqDecider(A)].  \mforall{}[eq':EqDecider(A')].  \mforall{}[f,g:a:A  fp->  B[a]].
        \{f  \msubseteq{}  g  supposing  f  \msubseteq{}  g\}  supposing  \mforall{}a:A.  (B[a]  \msubseteq{}r  C[a]) 
    supposing  strong-subtype(A;A')



Date html generated: 2015_07_17-AM-09_17_32
Last ObjectModification: 2015_01_28-AM-07_51_02

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