Nuprl Lemma : rless-cases-proof
∀x,y:ℝ. ((x < y) ⇒ (∀z:ℝ. ((x < z) ∨ (z < y))))
Proof
Definitions occuring in Statement :
rless: x < y,
real: ℝ,
all: ∀x:A. B[x],
implies: P ⇒ Q,
or: P ∨ Q
Definitions unfolded in proof :
all: ∀x:A. B[x],
implies: P ⇒ Q,
member: t ∈ T,
iff: P ⇐⇒ Q,
and: P ∧ Q,
sq_exists: ∃x:A [B[x]],
nat_plus: ℕ+,
uall: ∀[x:A]. B[x],
so_lambda: λ2x.t[x],
so_apply: x[s],
uimplies: b supposing a,
real: ℝ,
rless: x < y,
decidable: Dec(P),
or: P ∨ Q,
not: ¬A,
satisfiable_int_formula: satisfiable_int_formula(fmla),
exists: ∃x:A. B[x],
false: False,
top: Top,
prop: ℙ,
sq_type: SQType(T),
guard: {T},
less_than: a < b,
squash: ↓T,
uiff: uiff(P;Q)
Latex:
\mforall{}x,y:\mBbbR{}. ((x < y) {}\mRightarrow{} (\mforall{}z:\mBbbR{}. ((x < z) \mvee{} (z < y))))
Date html generated:
2020_05_20-AM-11_05_24
Last ObjectModification:
2019_11_25-PM-00_25_02
Theory : reals
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