Nuprl Lemma : rmaximum-lub
∀n,m:ℤ. ∀x:{n..m + 1-} ⟶ ℝ. ∀r:ℝ. (∀k:{n..m + 1-}. (x[k] ≤ r)) ⇒ (rmaximum(n;m;i.x[i]) ≤ r) supposing n ≤ m
Proof
Definitions occuring in Statement :
rmaximum: rmaximum(n;m;k.x[k]),
rleq: x ≤ y,
real: ℝ,
int_seg: {i..j-},
uimplies: b supposing a,
so_apply: x[s],
le: A ≤ B,
all: ∀x:A. B[x],
implies: P ⇒ Q,
function: x:A ⟶ B[x],
add: n + m,
natural_number: $n,
int: ℤ
Definitions unfolded in proof :
rmaximum: rmaximum(n;m;k.x[k]),
all: ∀x:A. B[x],
uimplies: b supposing a,
member: t ∈ T,
implies: P ⇒ Q,
uall: ∀[x:A]. B[x],
nat: ℕ,
false: False,
ge: i ≥ j ,
not: ¬A,
satisfiable_int_formula: satisfiable_int_formula(fmla),
exists: ∃x:A. B[x],
and: P ∧ Q,
prop: ℙ,
rleq: x ≤ y,
rnonneg: rnonneg(x),
le: A ≤ B,
bool: 𝔹,
unit: Unit,
it: ⋅,
btrue: tt,
uiff: uiff(P;Q),
ifthenelse: if b then t else f fi ,
rev_uimplies: rev_uimplies(P;Q),
so_apply: x[s],
int_seg: {i..j-},
lelt: i ≤ j < k,
decidable: Dec(P),
or: P ∨ Q,
rge: x ≥ y,
guard: {T},
bfalse: ff,
sq_type: SQType(T),
bnot: ¬bb,
assert: ↑b,
rev_implies: P ⇐ Q,
iff: P ⇐⇒ Q,
less_than: a < b,
squash: ↓T,
cand: A c∧ B,
req_int_terms: t1 ≡ t2,
subtype_rel: A ⊆r B,
subtract: n - m,
less_than': less_than'(a;b),
true: True
Latex:
\mforall{}n,m:\mBbbZ{}. \mforall{}x:\{n..m + 1\msupminus{}\} {}\mrightarrow{} \mBbbR{}. \mforall{}r:\mBbbR{}.
(\mforall{}k:\{n..m + 1\msupminus{}\}. (x[k] \mleq{} r)) {}\mRightarrow{} (rmaximum(n;m;i.x[i]) \mleq{} r) supposing n \mleq{} m
Date html generated:
2020_05_20-AM-11_14_25
Last ObjectModification:
2020_03_19-PM-05_56_48
Theory : reals
Home
Index