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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