Nuprl Lemma : subtype_rel_sets_simple
∀[A:Type]. ∀[P,Q:A ⟶ ℙ]. {a:A| P[a]} ⊆r {b:A| Q[b]} supposing ∀a:A. (P[a] ⇒ Q[a])
Proof
Definitions occuring in Statement :
uimplies: b supposing a,
subtype_rel: A ⊆r B,
uall: ∀[x:A]. B[x],
prop: ℙ,
so_apply: x[s],
all: ∀x:A. B[x],
implies: P ⇒ Q,
set: {x:A| B[x]} ,
function: x:A ⟶ B[x],
universe: Type
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
uimplies: b supposing a,
so_lambda: λ2x.t[x],
so_apply: x[s],
subtype_rel: A ⊆r B,
prop: ℙ,
all: ∀x:A. B[x],
implies: P ⇒ Q
Lemmas referenced :
subtype_rel_sets,
istype-universe
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
Error :isect_memberFormation_alt,
introduction,
cut,
extract_by_obid,
sqequalHypSubstitution,
isectElimination,
thin,
hypothesisEquality,
because_Cache,
sqequalRule,
Error :lambdaEquality_alt,
applyEquality,
Error :universeIsType,
independent_isectElimination,
setElimination,
rename,
hypothesis,
Error :setIsType,
universeEquality,
axiomEquality,
Error :functionIsType,
Error :isect_memberEquality_alt,
Error :isectIsTypeImplies,
Error :inhabitedIsType,
instantiate
Latex:
\mforall{}[A:Type]. \mforall{}[P,Q:A {}\mrightarrow{} \mBbbP{}]. \{a:A| P[a]\} \msubseteq{}r \{b:A| Q[b]\} supposing \mforall{}a:A. (P[a] {}\mRightarrow{} Q[a])
Date html generated:
2019_06_20-AM-11_19_27
Last ObjectModification:
2018_10_31-PM-03_32_44
Theory : subtype_0
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