Nuprl Lemma : per-isect2_quotient
∀[T:Type]. ∀[E1,E2:T ⟶ T ⟶ ℙ].
  (x,y:T/per/E1[x;y] ⋂ x,y:T/per/E2[x;y] ≡ x,y:T/per/(E1[x;y] ∧ E2[x;y])) supposing 
     (EquivRel(T;x,y.E1[x;y]) and 
     EquivRel(T;x,y.E2[x;y]))
Proof
Definitions occuring in Statement : 
per-quotient: x,y:T/per/E[x; y]
, 
equiv_rel: EquivRel(T;x,y.E[x; y])
, 
isect2: T1 ⋂ T2
, 
ext-eq: A ≡ B
, 
uimplies: b supposing a
, 
uall: ∀[x:A]. B[x]
, 
prop: ℙ
, 
so_apply: x[s1;s2]
, 
and: P ∧ Q
, 
function: x:A ⟶ B[x]
, 
universe: Type
Definitions unfolded in proof : 
per-quotient: x,y:T/per/E[x; y]
, 
quotient: x,y:A//B[x; y]
Lemmas referenced : 
isect2_quotient
Rules used in proof : 
cut, 
introduction, 
extract_by_obid, 
sqequalRule, 
sqequalReflexivity, 
sqequalSubstitution, 
sqequalTransitivity, 
computationStep, 
hypothesis
Latex:
\mforall{}[T:Type].  \mforall{}[E1,E2:T  {}\mrightarrow{}  T  {}\mrightarrow{}  \mBbbP{}].
    (x,y:T/per/E1[x;y]  \mcap{}  x,y:T/per/E2[x;y]  \mequiv{}  x,y:T/per/(E1[x;y]  \mwedge{}  E2[x;y]))  supposing 
          (EquivRel(T;x,y.E1[x;y])  and 
          EquivRel(T;x,y.E2[x;y]))
Date html generated:
2019_06_20-PM-00_33_39
Last ObjectModification:
2018_08_21-PM-10_54_13
Theory : per-quotient
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