Nuprl Lemma : fpf-sub-val2

[A,A':Type].
  ∀[B:A ─→ Type]
    ∀eq:EqDecider(A'). ∀f,g:a:A fp-> B[a]. ∀x:A'.
      ∀[P,Q:a:A ─→ B[a] ─→ ℙ].
        ((∀x:A. ∀z:B[x].  (P[x;z]  Q[x;z]))  != f(x) ==> P[x;z]  != g(x) ==> Q[x;z] supposing g ⊆ f) 
  supposing strong-subtype(A;A')


Proof




Definitions occuring in Statement :  fpf-sub: f ⊆ g fpf-val: != f(x) ==> P[a; z] fpf: a:A fp-> B[a] deq: EqDecider(T) strong-subtype: strong-subtype(A;B) uimplies: supposing a uall: [x:A]. B[x] prop: so_apply: x[s1;s2] so_apply: x[s] all: x:A. B[x] implies:  Q function: x:A ─→ B[x] universe: Type
Lemmas :  strong-subtype_witness strong-subtype-deq-subtype fpf-sub_witness fpf_ap_pair_lemma fpf-sub_wf all_wf fpf_wf deq_wf strong-subtype_wf assert-deq-member subtype_rel_list strong-subtype-l_member-type assert_wf deq-member_wf l_member_wf
\mforall{}[A,A':Type].
    \mforall{}[B:A  {}\mrightarrow{}  Type]
        \mforall{}eq:EqDecider(A').  \mforall{}f,g:a:A  fp->  B[a].  \mforall{}x:A'.
            \mforall{}[P,Q:a:A  {}\mrightarrow{}  B[a]  {}\mrightarrow{}  \mBbbP{}].
                ((\mforall{}x:A.  \mforall{}z:B[x].    (P[x;z]  {}\mRightarrow{}  Q[x;z]))
                {}\mRightarrow{}  z  !=  f(x)  ==>  P[x;z]  {}\mRightarrow{}  z  !=  g(x)  ==>  Q[x;z]  supposing  g  \msubseteq{}  f) 
    supposing  strong-subtype(A;A')



Date html generated: 2015_07_17-AM-11_08_02
Last ObjectModification: 2015_01_28-AM-07_53_59

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