Nuprl Lemma : ftc-integral
∀I:Interval
(iproper(I)
⇒ (∀a,b:{a:ℝ| a ∈ I} . ∀f:{f:I ⟶ℝ| ∀x,y:{a:ℝ| a ∈ I} . ((x = y)
⇒ ((f x) = (f y)))} . ∀g:I ⟶ℝ.
(d(g[x])/dx = λx.f[x] on I
⇒ (a_∫-b f[t] dt = (g[b] - g[a])))))
Proof
Definitions occuring in Statement :
integral: a_∫-b f[x] dx
,
derivative: d(f[x])/dx = λz.g[z] on I
,
rfun: I ⟶ℝ
,
i-member: r ∈ I
,
iproper: iproper(I)
,
interval: Interval
,
rsub: x - y
,
req: x = y
,
real: ℝ
,
so_apply: x[s]
,
all: ∀x:A. B[x]
,
implies: P
⇒ Q
,
set: {x:A| B[x]}
,
apply: f a
Definitions unfolded in proof :
all: ∀x:A. B[x]
,
member: t ∈ T
,
implies: P
⇒ Q
,
exists: ∃x:A. B[x]
,
prop: ℙ
,
uall: ∀[x:A]. B[x]
,
so_lambda: λ2x.t[x]
,
label: ...$L... t
,
rfun: I ⟶ℝ
,
so_apply: x[s]
,
integrate: a_∫- f[t] dt
,
subtype_rel: A ⊆r B
,
uiff: uiff(P;Q)
,
and: P ∧ Q
,
rev_uimplies: rev_uimplies(P;Q)
,
uimplies: b supposing a
,
rsub: x - y
,
guard: {T}
Lemmas referenced :
fund-theorem-of-calculus,
derivative_wf,
i-member_wf,
real_wf,
rfun_wf,
set_wf,
all_wf,
req_wf,
iproper_wf,
interval_wf,
integrate_wf,
radd_wf,
rsub_wf,
radd-preserves-req,
rminus_wf,
int-to-real_wf,
req_functionality,
req_weakening,
uiff_transitivity,
req_inversion,
radd-assoc,
radd_functionality,
radd-ac,
req_transitivity,
radd_comm,
radd-rminus-assoc,
radd-rminus-both,
integral-same-endpoints
Rules used in proof :
cut,
introduction,
extract_by_obid,
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
lambdaFormation,
hypothesis,
sqequalHypSubstitution,
dependent_functionElimination,
thin,
hypothesisEquality,
independent_functionElimination,
productElimination,
isectElimination,
sqequalRule,
lambdaEquality,
applyEquality,
setElimination,
rename,
dependent_set_memberEquality,
setEquality,
because_Cache,
functionEquality,
independent_isectElimination,
natural_numberEquality
Latex:
\mforall{}I:Interval
(iproper(I)
{}\mRightarrow{} (\mforall{}a,b:\{a:\mBbbR{}| a \mmember{} I\} . \mforall{}f:\{f:I {}\mrightarrow{}\mBbbR{}| \mforall{}x,y:\{a:\mBbbR{}| a \mmember{} I\} . ((x = y) {}\mRightarrow{} ((f x) = (f y)))\} . \mforall{}g:I {}\mrightarrow{}\mBbbR{}\000C.
(d(g[x])/dx = \mlambda{}x.f[x] on I {}\mRightarrow{} (a\_\mint{}\msupminus{}b f[t] dt = (g[b] - g[a])))))
Date html generated:
2016_10_26-PM-00_11_25
Last ObjectModification:
2016_09_12-PM-05_39_15
Theory : reals_2
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