Nuprl Lemma : itersetfun_functionality

G:Set{i:l} ⟶ Set{i:l}
  ((∀a,b:Set{i:l}.  ((a ⊆ b)  (G[a] ⊆ G[b])))
   (∀b,a:Set{i:l}.  (seteq(a;b)  seteq(itersetfun(x.G[x];a);itersetfun(x.G[x];b)))))


Proof




Definitions occuring in Statement :  itersetfun: itersetfun(s.G[s];a) setsubset: (a ⊆ b) seteq: seteq(s1;s2) Set: Set{i:l} so_apply: x[s] all: x:A. B[x] implies:  Q function: x:A ⟶ B[x]
Definitions unfolded in proof :  prop: cand: c∧ B rev_implies:  Q so_apply: x[s] so_lambda: λ2x.t[x] uall: [x:A]. B[x] and: P ∧ Q iff: ⇐⇒ Q member: t ∈ T implies:  Q all: x:A. B[x]
Lemmas referenced :  setsubset_wf all_wf seteq_wf itersetfun_functionality_subset Set_wf itersetfun_wf seteq-iff-setsubset
Rules used in proof :  functionEquality cumulativity instantiate because_Cache independent_pairFormation applyEquality lambdaEquality sqequalRule isectElimination independent_functionElimination productElimination hypothesis hypothesisEquality thin dependent_functionElimination sqequalHypSubstitution extract_by_obid introduction cut lambdaFormation sqequalReflexivity computationStep sqequalTransitivity sqequalSubstitution

Latex:
\mforall{}G:Set\{i:l\}  {}\mrightarrow{}  Set\{i:l\}
    ((\mforall{}a,b:Set\{i:l\}.    ((a  \msubseteq{}  b)  {}\mRightarrow{}  (G[a]  \msubseteq{}  G[b])))
    {}\mRightarrow{}  (\mforall{}b,a:Set\{i:l\}.    (seteq(a;b)  {}\mRightarrow{}  seteq(itersetfun(x.G[x];a);itersetfun(x.G[x];b)))))



Date html generated: 2018_05_23-AM-08_10_11
Last ObjectModification: 2018_05_22-PM-11_28_45

Theory : constructive!set!theory


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