Nuprl Lemma : fpf-join-ap
∀[A:Type]. ∀[B:A ─→ Type]. ∀[eq:EqDecider(A)]. ∀[f,g:a:A fp-> B[a]]. ∀[x:A].
f ⊕ g(x) = if x ∈ dom(f) then f(x) else g(x) fi ∈ B[x] supposing ↑x ∈ dom(f ⊕ g)
Proof
Definitions occuring in Statement :
fpf-join: f ⊕ g,
fpf-ap: f(x),
fpf-dom: x ∈ dom(f),
fpf: a:A fp-> B[a],
deq: EqDecider(T),
assert: ↑b,
ifthenelse: if b then t else f fi ,
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
so_apply: x[s],
function: x:A ─→ B[x],
universe: Type,
equal: s = t ∈ T
Lemmas :
fpf-ap_wf,
fpf-join_wf,
assert_wf,
fpf-dom_wf,
top_wf,
subtype-fpf2,
subtype_top,
deq_wf
\mforall{}[A:Type]. \mforall{}[B:A {}\mrightarrow{} Type]. \mforall{}[eq:EqDecider(A)]. \mforall{}[f,g:a:A fp-> B[a]]. \mforall{}[x:A].
f \moplus{} g(x) = if x \mmember{} dom(f) then f(x) else g(x) fi supposing \muparrow{}x \mmember{} dom(f \moplus{} g)
Date html generated:
2015_07_17-AM-09_19_59
Last ObjectModification:
2015_01_28-AM-07_48_54
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