Nuprl Lemma : fpf-sub-val3
∀[A:Type]. ∀[B,C:A ⟶ Type].
∀eq:EqDecider(A). ∀f:a:A fp-> B[a]. ∀g:a:A fp-> C[a]. ∀x:A.
∀[P:a:A ⟶ B[a] ⟶ ℙ]. ∀[Q:a:A ⟶ C[a] ⟶ ℙ].
((∀x:A. ((C[x] ⊆r B[x]) c∧ (P[x;g(x)] ⇒ Q[x;g(x)])) supposing ((↑x ∈ dom(f)) and (↑x ∈ dom(g))))
⇒ {z != f(x) ==> P[y;z] ⇒ z != g(x) ==> Q[y;z] supposing g ⊆ f})
Proof
Definitions occuring in Statement :
fpf-sub: f ⊆ g,
fpf-val: z != f(x) ==> P[a; z],
fpf-ap: f(x),
fpf-dom: x ∈ dom(f),
fpf: a:A fp-> B[a],
deq: EqDecider(T),
assert: ↑b,
uimplies: b supposing a,
subtype_rel: A ⊆r B,
uall: ∀[x:A]. B[x],
cand: A c∧ B,
prop: ℙ,
guard: {T},
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,
guard: {T},
uall: ∀[x:A]. B[x],
all: ∀x:A. B[x],
implies: P ⇒ Q,
uimplies: b supposing a,
member: t ∈ T,
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,C:A {}\mrightarrow{} Type].
\mforall{}eq:EqDecider(A). \mforall{}f:a:A fp-> B[a]. \mforall{}g:a:A fp-> C[a]. \mforall{}x:A.
\mforall{}[P:a:A {}\mrightarrow{} B[a] {}\mrightarrow{} \mBbbP{}]. \mforall{}[Q:a:A {}\mrightarrow{} C[a] {}\mrightarrow{} \mBbbP{}].
((\mforall{}x:A
((C[x] \msubseteq{}r B[x]) c\mwedge{} (P[x;g(x)] {}\mRightarrow{} Q[x;g(x)])) supposing ((\muparrow{}x \mmember{} dom(f)) and (\muparrow{}x \mmember{} dom(g))))
{}\mRightarrow{} \{z != f(x) ==> P[y;z] {}\mRightarrow{} z != g(x) ==> Q[y;z] supposing g \msubseteq{} f\})
Date html generated:
2016_05_16-AM-11_15_23
Last ObjectModification:
2015_12_29-AM-09_20_00
Theory : event-ordering
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