Nuprl Lemma : rfun-ap_functionality
∀[f:ℝ ⟶ ℝ]. ∀x,y:ℝ. f(x) = f(y) supposing x = y supposing ∀x,y:ℝ. ((x = y) ⇒ ((f x) = (f y)))
Proof
Definitions occuring in Statement :
rfun-ap: f(x),
req: x = y,
real: ℝ,
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
all: ∀x:A. B[x],
implies: P ⇒ Q,
apply: f a,
function: x:A ⟶ B[x]
Definitions unfolded in proof :
rfun-ap: f(x),
uall: ∀[x:A]. B[x],
member: t ∈ T,
uimplies: b supposing a,
all: ∀x:A. B[x],
implies: P ⇒ Q,
prop: ℙ,
so_lambda: λ2x.t[x],
so_apply: x[s],
guard: {T}
Lemmas referenced :
req_witness,
real_wf,
req_wf,
all_wf
Rules used in proof :
sqequalSubstitution,
sqequalRule,
sqequalReflexivity,
sqequalTransitivity,
computationStep,
isect_memberFormation,
introduction,
cut,
lambdaFormation,
extract_by_obid,
sqequalHypSubstitution,
isectElimination,
thin,
applyEquality,
functionExtensionality,
hypothesisEquality,
hypothesis,
independent_functionElimination,
lambdaEquality,
dependent_functionElimination,
isect_memberEquality,
because_Cache,
equalityTransitivity,
equalitySymmetry,
functionEquality
Latex:
\mforall{}[f:\mBbbR{} {}\mrightarrow{} \mBbbR{}]. \mforall{}x,y:\mBbbR{}. f(x) = f(y) supposing x = y supposing \mforall{}x,y:\mBbbR{}. ((x = y) {}\mRightarrow{} ((f x) = (f y)))
Date html generated:
2017_10_04-PM-11_02_13
Last ObjectModification:
2017_06_30-PM-03_20_36
Theory : reals_2
Home
Index