Nuprl Lemma : fpf-sub-val2
∀[A,A':Type].
∀[B:A ⟶ Type]
∀eq:EqDecider(A'). ∀f,g:a:A fp-> B[a]. ∀x:A'.
∀[P,Q:a:A ⟶ B[a] ⟶ ℙ].
((∀x:A. ∀z:B[x]. (P[x;z] ⇒ Q[x;z])) ⇒ z != f(x) ==> P[x;z] ⇒ z != g(x) ==> Q[x;z] supposing g ⊆ f)
supposing strong-subtype(A;A')
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),
strong-subtype: strong-subtype(A;B),
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 :
uall: ∀[x:A]. B[x],
uimplies: b supposing a,
member: t ∈ T,
implies: P ⇒ Q,
all: ∀x:A. B[x],
so_lambda: λ2x.t[x],
so_apply: x[s],
subtype_rel: A ⊆r B,
prop: ℙ,
so_apply: x[s1;s2],
fpf-val: z != f(x) ==> P[a; z],
fpf: a:A fp-> B[a],
fpf-dom: x ∈ dom(f),
pi1: fst(t),
top: Top,
fpf-sub: f ⊆ g,
deq-member: x ∈b L,
reduce: reduce(f;k;as),
list_ind: list_ind,
strong-subtype: strong-subtype(A;B),
cand: A c∧ B,
iff: P ⇐⇒ Q,
and: P ∧ Q,
rev_implies: P ⇐ Q,
fpf-ap: f(x),
pi2: snd(t)
Latex:
\mforall{}[A,A':Type].
\mforall{}[B:A {}\mrightarrow{} Type]
\mforall{}eq:EqDecider(A'). \mforall{}f,g:a:A fp-> B[a]. \mforall{}x:A'.
\mforall{}[P,Q:a:A {}\mrightarrow{} B[a] {}\mrightarrow{} \mBbbP{}].
((\mforall{}x:A. \mforall{}z:B[x]. (P[x;z] {}\mRightarrow{} Q[x;z]))
{}\mRightarrow{} z != f(x) ==> P[x;z] {}\mRightarrow{} z != g(x) ==> Q[x;z] supposing g \msubseteq{} f)
supposing strong-subtype(A;A')
Date html generated:
2016_05_16-AM-11_15_19
Last ObjectModification:
2015_12_29-AM-09_24_43
Theory : event-ordering
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