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
Lemmas :
top_wf,
subtype_base_sq,
fpf_wf,
fpf-dom_wf,
assert_wf,
subtype_top,
subtype-fpf2,
cand_wf,
all_wf,
fpf-ap_wf,
assert_witness,
deq_wf,
equal-wf-base,
equal-wf-base-T,
equal-wf-T-base
\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:
2015_07_17-AM-11_07_52
Last ObjectModification:
2015_01_28-AM-07_47_00
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