Nuprl Lemma : mapfilter-subbag
∀T,U:Type. ∀f:T ⟶ U. ∀P,Q:T ⟶ 𝔹. ∀L:T List.
((∀t:T. ((↑(P t)) ⇒ (↑(Q t)))) ⇒ sub-bag(U;mapfilter(f;P;L);mapfilter(f;Q;L)))
Proof
Definitions occuring in Statement :
mapfilter: mapfilter(f;P;L),
list: T List,
assert: ↑b,
bool: 𝔹,
all: ∀x:A. B[x],
implies: P ⇒ Q,
apply: f a,
function: x:A ⟶ B[x],
universe: Type,
sub-bag: sub-bag(T;as;bs)
Definitions unfolded in proof :
all: ∀x:A. B[x],
implies: P ⇒ Q,
sub-bag: sub-bag(T;as;bs),
exists: ∃x:A. B[x],
member: t ∈ T,
uall: ∀[x:A]. B[x],
so_apply: x[s],
bool: 𝔹,
unit: Unit,
it: ⋅,
btrue: tt,
band: p ∧b q,
ifthenelse: if b then t else f fi ,
uiff: uiff(P;Q),
and: P ∧ Q,
uimplies: b supposing a,
prop: ℙ,
or: P ∨ Q,
sq_type: SQType(T),
guard: {T},
bnot: ¬bb,
assert: ↑b,
bfalse: ff,
false: False,
subtype_rel: A ⊆r B,
so_lambda: λ2x.t[x],
bag-append: as + bs,
bor: p ∨bq,
iff: P ⇐⇒ Q,
rev_implies: P ⇐ Q,
true: True,
not: ¬A,
top: Top,
l_all: (∀x∈L.P[x])
Latex:
\mforall{}T,U:Type. \mforall{}f:T {}\mrightarrow{} U. \mforall{}P,Q:T {}\mrightarrow{} \mBbbB{}. \mforall{}L:T List.
((\mforall{}t:T. ((\muparrow{}(P t)) {}\mRightarrow{} (\muparrow{}(Q t)))) {}\mRightarrow{} sub-bag(U;mapfilter(f;P;L);mapfilter(f;Q;L)))
Date html generated:
2016_05_17-AM-08_56_26
Last ObjectModification:
2015_12_29-PM-02_55_25
Theory : messages
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