Nuprl Lemma : pair_eta_rw

`∀[A:Type]. ∀[B:A ⟶ Type]. ∀[p:a:A × B[a]].  (<fst(p), snd(p)> = p ∈ (a:A × B[a]))`

Proof

Definitions occuring in Statement :  uall: `∀[x:A]. B[x]` so_apply: `x[s]` pi1: `fst(t)` pi2: `snd(t)` function: `x:A ⟶ B[x]` pair: `<a, b>` product: `x:A × B[x]` universe: `Type` equal: `s = t ∈ T`
Definitions unfolded in proof :  uall: `∀[x:A]. B[x]` member: `t ∈ T` so_apply: `x[s]` pi1: `fst(t)` pi2: `snd(t)`
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity Error :isect_memberFormation_alt,  introduction cut hypothesis Error :productIsType,  Error :universeIsType,  hypothesisEquality applyEquality sqequalRule sqequalHypSubstitution isect_memberEquality isectElimination thin axiomEquality productEquality Error :functionIsType,  Error :inhabitedIsType,  because_Cache functionEquality cumulativity universeEquality productElimination dependent_pairEquality

Latex:
\mforall{}[A:Type].  \mforall{}[B:A  {}\mrightarrow{}  Type].  \mforall{}[p:a:A  \mtimes{}  B[a]].    (<fst(p),  snd(p)>  =  p)

Date html generated: 2019_06_20-AM-11_14_42
Last ObjectModification: 2018_09_26-AM-10_42_03

Theory : core_2

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