Nuprl Lemma : rmax_strict_ub
∀x,y,z:ℝ. ((z < x) ∨ (z < y) ⇐⇒ z < rmax(x;y))
Proof
Definitions occuring in Statement :
rless: x < y,
rmax: rmax(x;y),
real: ℝ,
all: ∀x:A. B[x],
iff: P ⇐⇒ Q,
or: P ∨ Q
Definitions unfolded in proof :
all: ∀x:A. B[x],
iff: P ⇐⇒ Q,
and: P ∧ Q,
implies: P ⇒ Q,
or: P ∨ Q,
member: t ∈ T,
uall: ∀[x:A]. B[x],
prop: ℙ,
rev_implies: P ⇐ Q,
cand: A c∧ B,
guard: {T},
uimplies: b supposing a,
exists: ∃x:A. B[x],
rmax: rmax(x;y),
squash: ↓T,
real: ℝ,
true: True,
subtype_rel: A ⊆r B,
not: ¬A,
false: False,
sq_type: SQType(T),
uiff: uiff(P;Q),
ifthenelse: if b then t else f fi ,
btrue: tt,
bfalse: ff
Latex:
\mforall{}x,y,z:\mBbbR{}. ((z < x) \mvee{} (z < y) \mLeftarrow{}{}\mRightarrow{} z < rmax(x;y))
Date html generated:
2020_05_20-AM-10_58_24
Last ObjectModification:
2019_11_11-PM-09_03_41
Theory : reals
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