Nuprl Lemma : empty-bag-union
∀[T:Type]. ∀[bbs:bag(bag(T))]. ∀bs:bag(T). (bs ↓∈ bbs ⇒ (bs = {} ∈ bag(T))) supposing bag-union(bbs) = {} ∈ bag(T)
Proof
Definitions occuring in Statement :
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
all: ∀x:A. B[x],
implies: P ⇒ Q,
universe: Type,
equal: s = t ∈ T,
bag-member: x ↓∈ bs,
bag-union: bag-union(bbs),
empty-bag: {},
bag: bag(T)
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
uimplies: b supposing a,
all: ∀x:A. B[x],
implies: P ⇒ Q,
prop: ℙ,
exists: ∃x:A. B[x],
subtype_rel: A ⊆r B,
nat: ℕ,
sq_type: SQType(T),
guard: {T},
squash: ↓T,
true: True,
iff: P ⇐⇒ Q,
and: P ∧ Q,
rev_implies: P ⇐ Q
Latex:
\mforall{}[T:Type]. \mforall{}[bbs:bag(bag(T))]. \mforall{}bs:bag(T). (bs \mdownarrow{}\mmember{} bbs {}\mRightarrow{} (bs = \{\})) supposing bag-union(bbs) = \{\}
Date html generated:
2016_05_17-AM-11_10_04
Last ObjectModification:
2016_01_18-AM-00_11_14
Theory : process-model
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