Nuprl Lemma : fpf-sub-val
∀[A:Type]. ∀[B:A ⟶ Type].
∀eq:EqDecider(A). ∀f,g:a:A fp-> B[a]. ∀x:A.
∀[P:a:A ⟶ B[a] ⟶ ℙ]. z != f(x) ==> P[x;z] ⇒ z != g(x) ==> P[x;z] supposing g ⊆ f
Proof
Definitions occuring in Statement :
fpf-sub: f ⊆ g,
fpf-val: z != f(x) ==> P[a; z],
fpf: a:A fp-> B[a],
deq: EqDecider(T),
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
prop: ℙ,
so_apply: x[s1;s2],
so_apply: x[s],
all: ∀x:A. B[x],
implies: P ⇒ Q,
function: x:A ⟶ B[x],
universe: Type
Definitions unfolded in proof :
fpf-val: z != f(x) ==> P[a; z],
fpf-sub: f ⊆ g,
uall: ∀[x:A]. B[x],
all: ∀x:A. B[x],
uimplies: b supposing a,
member: t ∈ T,
implies: P ⇒ Q,
cand: A c∧ B,
subtype_rel: A ⊆r B,
so_lambda: λ2x.t[x],
so_apply: x[s],
top: Top,
prop: ℙ,
so_apply: x[s1;s2]
Latex:
\mforall{}[A:Type]. \mforall{}[B:A {}\mrightarrow{} Type].
\mforall{}eq:EqDecider(A). \mforall{}f,g:a:A fp-> B[a]. \mforall{}x:A.
\mforall{}[P:a:A {}\mrightarrow{} B[a] {}\mrightarrow{} \mBbbP{}]. z != f(x) ==> P[x;z] {}\mRightarrow{} z != g(x) ==> P[x;z] supposing g \msubseteq{} f
Date html generated:
2016_05_16-AM-11_15_05
Last ObjectModification:
2015_12_29-AM-09_19_42
Theory : event-ordering
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