Nuprl Lemma : fpf-sub-join-right2

[A:Type]. ∀[B,C:A ─→ Type]. ∀[eq:EqDecider(A)]. ∀[f:a:A fp-> B[a]]. ∀[g:a:A fp-> C[a]].
  g ⊆ f ⊕ supposing ∀x:A. (((↑x ∈ dom(f)) ∧ (↑x ∈ dom(g)))  ((B[x] ⊆C[x]) c∧ (f(x) g(x) ∈ C[x])))


Proof




Definitions occuring in Statement :  fpf-join: f ⊕ g fpf-sub: f ⊆ g fpf-ap: f(x) fpf-dom: x ∈ dom(f) fpf: a:A fp-> B[a] deq: EqDecider(T) assert: b uimplies: supposing a subtype_rel: A ⊆B uall: [x:A]. B[x] cand: c∧ B so_apply: x[s] all: x:A. B[x] implies:  Q and: P ∧ Q function: x:A ─→ B[x] universe: Type equal: t ∈ T
Lemmas :  assert_witness fpf-dom_wf fpf-join_wf top_wf assert_wf all_wf subtype-fpf2 subtype_top subtype_rel_wf fpf-ap_wf fpf_wf deq_wf fpf-join-dom2 bool_wf equal-wf-T-base bnot_wf not_wf eqtt_to_assert uiff_transitivity eqff_to_assert assert_of_bnot
\mforall{}[A:Type].  \mforall{}[B,C:A  {}\mrightarrow{}  Type].  \mforall{}[eq:EqDecider(A)].  \mforall{}[f:a:A  fp->  B[a]].  \mforall{}[g:a:A  fp->  C[a]].
    g  \msubseteq{}  f  \moplus{}  g  supposing  \mforall{}x:A.  (((\muparrow{}x  \mmember{}  dom(f))  \mwedge{}  (\muparrow{}x  \mmember{}  dom(g)))  {}\mRightarrow{}  ((B[x]  \msubseteq{}r  C[x])  c\mwedge{}  (f(x)  =  g(x))))



Date html generated: 2015_07_17-AM-09_20_35
Last ObjectModification: 2015_01_28-AM-07_49_29

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