Nuprl Lemma : opposite-lattice-0
∀[L:Top]. (0 ~ 1)
Proof
Definitions occuring in Statement :
opposite-lattice: opposite-lattice(L),
lattice-0: 0,
lattice-1: 1,
uall: ∀[x:A]. B[x],
top: Top,
sqequal: s ~ t
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
opposite-lattice: opposite-lattice(L),
lattice-0: 0,
so_lambda: λ2x y.t[x; y],
mk-bounded-distributive-lattice: mk-bounded-distributive-lattice,
mk-bounded-lattice: mk-bounded-lattice(T;m;j;z;o),
all: ∀x:A. B[x],
top: Top,
eq_atom: x =a y,
ifthenelse: if b then t else f fi ,
bfalse: ff,
btrue: tt
Lemmas referenced :
rec_select_update_lemma,
top_wf
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
isect_memberFormation,
introduction,
cut,
sqequalRule,
lemma_by_obid,
sqequalHypSubstitution,
dependent_functionElimination,
thin,
isect_memberEquality,
voidElimination,
voidEquality,
hypothesis,
axiomSqEquality
Latex:
\mforall{}[L:Top]. (0 \msim{} 1)
Date html generated:
2020_05_20-AM-08_47_10
Last ObjectModification:
2015_12_28-PM-02_00_52
Theory : lattices
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