Nuprl Lemma : extensional-real-to-bool-constant
∀f:ℝ ⟶ 𝔹. ∀x,y:ℝ. f x = f y supposing ∀x,y:ℝ. ((x = y) ⇒ f x = f y)
Proof
Definitions occuring in Statement :
req: x = y,
real: ℝ,
bool: 𝔹,
uimplies: b supposing a,
all: ∀x:A. B[x],
implies: P ⇒ Q,
apply: f a,
function: x:A ⟶ B[x],
equal: s = t ∈ T
Definitions unfolded in proof :
all: ∀x:A. B[x],
uimplies: b supposing a,
member: t ∈ T,
decidable: Dec(P),
or: P ∨ Q,
not: ¬A,
implies: P ⇒ Q,
true: True,
false: False,
uall: ∀[x:A]. B[x],
prop: ℙ,
so_lambda: λ2x.t[x],
guard: {T},
sq_type: SQType(T),
cand: A c∧ B,
and: P ∧ Q,
top: Top,
bfalse: ff,
btrue: tt,
ifthenelse: if b then t else f fi ,
uiff: uiff(P;Q),
iff: P ⇐⇒ Q
Latex:
\mforall{}f:\mBbbR{} {}\mrightarrow{} \mBbbB{}. \mforall{}x,y:\mBbbR{}. f x = f y supposing \mforall{}x,y:\mBbbR{}. ((x = y) {}\mRightarrow{} f x = f y)
Date html generated:
2020_05_20-PM-00_05_12
Last ObjectModification:
2020_01_09-PM-06_15_14
Theory : reals
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