Nuprl Lemma : fpf-sub_functionality2
∀[A,A':Type].
∀[B:A ⟶ Type]. ∀[C:A' ⟶ Type]. ∀[eq:EqDecider(A)]. ∀[eq':EqDecider(A')]. ∀[f,g:a:A fp-> B[a]].
{f ⊆ g supposing f ⊆ g} supposing ∀a:A. (B[a] ⊆r C[a])
supposing strong-subtype(A;A')
Proof
Definitions occuring in Statement :
fpf-sub: f ⊆ g,
fpf: a:A fp-> B[a],
deq: EqDecider(T),
strong-subtype: strong-subtype(A;B),
uimplies: b supposing a,
subtype_rel: A ⊆r B,
uall: ∀[x:A]. B[x],
guard: {T},
so_apply: x[s],
all: ∀x:A. B[x],
function: x:A ⟶ B[x],
universe: Type
Definitions unfolded in proof :
guard: {T}
Lemmas referenced :
fpf-sub-functionality2
Rules used in proof :
cut,
lemma_by_obid,
sqequalSubstitution,
sqequalRule,
sqequalReflexivity,
sqequalTransitivity,
computationStep,
hypothesis
Latex:
\mforall{}[A,A':Type].
\mforall{}[B:A {}\mrightarrow{} Type]. \mforall{}[C:A' {}\mrightarrow{} Type]. \mforall{}[eq:EqDecider(A)]. \mforall{}[eq':EqDecider(A')]. \mforall{}[f,g:a:A fp-> B[a]].
\{f \msubseteq{} g supposing f \msubseteq{} g\} supposing \mforall{}a:A. (B[a] \msubseteq{}r C[a])
supposing strong-subtype(A;A')
Date html generated:
2018_05_21-PM-09_19_04
Last ObjectModification:
2018_02_09-AM-10_17_23
Theory : finite!partial!functions
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