Nuprl Lemma : sub-bag-member
∀[T:Type]. ∀[b1,b2:bag(T)]. ∀[x:T]. (x ↓∈ b2) supposing (sub-bag(T;b1;b2) and x ↓∈ b1)
Proof
Definitions occuring in Statement :
bag-member: x ↓∈ bs,
sub-bag: sub-bag(T;as;bs),
bag: bag(T),
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
universe: Type
Definitions unfolded in proof :
sub-bag: sub-bag(T;as;bs),
exists: ∃x:A. B[x],
uall: ∀[x:A]. B[x],
member: t ∈ T,
all: ∀x:A. B[x],
iff: P ⇐⇒ Q,
and: P ∧ Q,
rev_implies: P ⇐ Q,
implies: P ⇒ Q,
sq_or: a ↓∨ b,
or: P ∨ Q,
prop: ℙ,
squash: ↓T,
uimplies: b supposing a,
bag-member: x ↓∈ bs
Lemmas referenced :
bag-member-append,
bag-member_wf,
sub-bag_wf,
bag_wf
Rules used in proof :
sqequalHypSubstitution,
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
productElimination,
thin,
cut,
hypothesis,
introduction,
extract_by_obid,
isectElimination,
because_Cache,
dependent_functionElimination,
hypothesisEquality,
independent_functionElimination,
inlFormation,
cumulativity,
sqequalRule,
imageMemberEquality,
baseClosed,
hyp_replacement,
equalitySymmetry,
Error :applyLambdaEquality,
universeEquality,
isect_memberFormation,
imageElimination,
isect_memberEquality,
equalityTransitivity
Latex:
\mforall{}[T:Type]. \mforall{}[b1,b2:bag(T)]. \mforall{}[x:T]. (x \mdownarrow{}\mmember{} b2) supposing (sub-bag(T;b1;b2) and x \mdownarrow{}\mmember{} b1)
Date html generated:
2016_10_25-AM-10_30_23
Last ObjectModification:
2016_07_12-AM-06_46_39
Theory : bags
Home
Index