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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