Nuprl Lemma : s_part_char
∀[T:Type]. ∀[R:T ⟶ T ⟶ ℙ]. ∀[a,b:T]. (((R\) a b) = ((R a b) ∧ (¬(R b a))) ∈ ℙ)
Proof
Definitions occuring in Statement :
s_part: E\,
uall: ∀[x:A]. B[x],
prop: ℙ,
not: ¬A,
and: P ∧ Q,
apply: f a,
function: x:A ⟶ B[x],
universe: Type,
equal: s = t ∈ T
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
s_part: E\,
prop: ℙ
Lemmas referenced :
and_wf,
not_wf
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
isect_memberFormation,
introduction,
cut,
sqequalRule,
lemma_by_obid,
sqequalHypSubstitution,
isectElimination,
thin,
applyEquality,
hypothesisEquality,
hypothesis,
isect_memberEquality,
axiomEquality,
because_Cache,
functionEquality,
cumulativity,
universeEquality
Latex:
\mforall{}[T:Type]. \mforall{}[R:T {}\mrightarrow{} T {}\mrightarrow{} \mBbbP{}]. \mforall{}[a,b:T]. (((R\mbackslash{}) a b) = ((R a b) \mwedge{} (\mneg{}(R b a))))
Date html generated:
2016_05_15-PM-00_01_37
Last ObjectModification:
2015_12_26-PM-11_26_04
Theory : gen_algebra_1
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