Nuprl Lemma : type-with-x=0-y=1

∀x,y:Base.  ∃T:Type. ((x = 0 ∈ T) ∧ (y = 1 ∈ T) ∧ ((0 = 1 ∈ T) ⇒ (↓(x = y ∈ Base) ∨ (x = 1 ∈ Base) ∨ (y = 0 ∈ Base))))


Proof




Definitions occuring in Statement :  all: ∀x:A. B[x],  exists: ∃x:A. B[x],  squash: ↓T,  implies: P ⇒ Q,  or: P ∨ Q,  and: P ∧ Q,  natural_number: $n,  base: Base,  universe: Type,  equal: s = t ∈ T
Definitions unfolded in proof :  all: ∀x:A. B[x],  member: t ∈ T,  equiv_rel: EquivRel(T;x,y.E[x; y]),  trans: Trans(T;x,y.E[x; y]),  sym: Sym(T;x,y.E[x; y]),  refl: Refl(T;x,y.E[x; y]),  and: P ∧ Q,  cand: A c∧ B,  guard: {T},  or: P ∨ Q,  prop: ℙ,  uall: ∀[x:A]. B[x],  so_lambda: λ2x.t[x],  so_apply: x[s],  implies: P ⇒ Q,  uimplies: b supposing a,  sq_type: SQType(T),  true: True,  false: False,  exists: ∃x:A. B[x],  so_lambda: λ2x y.t[x; y],  so_apply: x[s1;s2],  squash: ↓T,  not: ¬A,  iff: P ⇐⇒ Q,  rev_implies: P ⇐ Q,  quotient: x,y:A//B[x; y]
Lemmas referenced :  base_wf,  or_wf,  equal-wf-base,  set_wf,  subtype_base_sq,  subtype_rel_self,  int_subtype_base,  quotient_wf,  quotient-member-eq,  squash_wf,  iff_weakening_equal
Rules used in proof :  sqequalSubstitution,  sqequalTransitivity,  computationStep,  sqequalReflexivity,  lambdaFormation,  cut,  introduction,  extract_by_obid,  hypothesis,  sqequalRule,  inrFormation,  inlFormation,  setElimination,  thin,  rename,  hypothesisEquality,  sqequalHypSubstitution,  isectElimination,  productEquality,  because_Cache,  baseClosed,  lambdaEquality,  independent_pairFormation,  unionElimination,  equalitySymmetry,  equalityTransitivity,  productElimination,  instantiate,  cumulativity,  independent_isectElimination,  dependent_functionElimination,  independent_functionElimination,  natural_numberEquality,  intEquality,  voidElimination,  promote_hyp,  dependent_pairFormation,  setEquality,  dependent_set_memberEquality,  imageElimination,  imageMemberEquality,  functionEquality,  sqequalIntensionalEquality,  pertypeElimination

Latex:
\mforall{}x,y:Base.    \mexists{}T:Type.  ((x  =  0)  \mwedge{}  (y  =  1)  \mwedge{}  ((0  =  1)  {}\mRightarrow{}  (\mdownarrow{}(x  =  y)  \mvee{}  (x  =  1)  \mvee{}  (y  =  0))))



Date html generated: 2018_05_21-PM-01_14_27
Last ObjectModification: 2018_05_01-PM-04_37_27

Theory : num_thy_1


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