Nuprl Lemma : sum-of-three-cubes-iff-1
This solution to the problem of writing 33 as the sum of three cubes
was found around March 9, 2019 by Andrew Booker using 15 core-years
computation time (over three weeks real time) on a super-computer in Bristol.
The smallest number for which it is unknown whether it is the sum of three
cubes is now 42 (and the next is 114).⋅
∀k:ℕ. ∀a,b,c:ℤ.
((0 ≤ (a + b))
⇒ ((((a * a * a) + (b * b * b)) + (c * c * c)) = k ∈ ℤ
⇐⇒ ∃d:ℕ
(((a + b) = d ∈ ℤ)
∧ (((d = 0 ∈ ℤ) ∧ ((c * c * c) = k ∈ ℤ))
∨ ((¬(d = 0 ∈ ℤ))
∧ ((k - c * c * c rem d) = 0 ∈ ℤ)
∧ (((a * a) + ((b * b) - a * b)) = ((k - c * c * c) ÷ d) ∈ ℤ))))))
Proof
Definitions occuring in Statement :
nat: ℕ,
le: A ≤ B,
all: ∀x:A. B[x],
exists: ∃x:A. B[x],
iff: P ⇐⇒ Q,
not: ¬A,
implies: P ⇒ Q,
or: P ∨ Q,
and: P ∧ Q,
remainder: n rem m,
divide: n ÷ m,
multiply: n * m,
subtract: n - m,
add: n + m,
natural_number: $n,
int: ℤ,
equal: s = t ∈ T
Definitions unfolded in proof :
uiff: uiff(P;Q),
cand: A c∧ B,
nequal: a ≠ b ∈ T ,
int_nzero: ℤ-o,
iff: P ⇐⇒ Q,
and: P ∧ Q,
subtype_rel: A ⊆r B,
so_lambda: λ2x.t[x],
so_apply: x[s],
rev_implies: P ⇐ Q,
guard: {T},
sq_type: SQType(T),
prop: ℙ,
top: Top,
false: False,
exists: ∃x:A. B[x],
satisfiable_int_formula: satisfiable_int_formula(fmla),
not: ¬A,
or: P ∨ Q,
decidable: Dec(P),
ge: i ≥ j ,
nat: ℕ,
uimplies: b supposing a,
member: t ∈ T,
uall: ∀[x:A]. B[x],
implies: P ⇒ Q,
all: ∀x:A. B[x]
Lemmas referenced :
decidable__le,
intformle_wf,
int_formula_prop_le_lemma,
div-cancel,
add-is-int-iff,
false_wf,
nequal_wf,
subtract_wf,
div_rem_sum,
itermConstant_wf,
int_term_value_constant_lemma,
intformand_wf,
int_formula_prop_and_lemma,
set_subtype_base,
le_wf,
int_formula_prop_wf,
int_term_value_subtract_lemma,
int_term_value_var_lemma,
int_term_value_mul_lemma,
int_term_value_add_lemma,
int_formula_prop_eq_lemma,
istype-void,
int_formula_prop_not_lemma,
itermSubtract_wf,
itermVar_wf,
itermMultiply_wf,
itermAdd_wf,
intformeq_wf,
intformnot_wf,
full-omega-unsat,
decidable__equal_int,
nat_properties,
int_subtype_base,
subtype_base_sq,
istype-nat,
istype-int,
istype-le
Rules used in proof :
pointwiseFunctionality,
promote_hyp,
Error :inrFormation_alt,
Error :dependent_set_memberEquality_alt,
independent_pairFormation,
Error :inlFormation_alt,
multiplyEquality,
Error :productIsType,
Error :functionIsType,
Error :equalityIstype,
Error :inhabitedIsType,
baseClosed,
sqequalBase,
baseApply,
closedConclusion,
applyEquality,
productElimination,
Error :unionIsType,
equalitySymmetry,
equalityTransitivity,
Error :universeIsType,
sqequalRule,
voidElimination,
Error :isect_memberEquality_alt,
int_eqEquality,
Error :lambdaEquality_alt,
Error :dependent_pairFormation_alt,
independent_functionElimination,
approximateComputation,
unionElimination,
dependent_functionElimination,
rename,
setElimination,
independent_isectElimination,
intEquality,
cumulativity,
instantiate,
because_Cache,
hypothesis,
hypothesisEquality,
addEquality,
natural_numberEquality,
thin,
isectElimination,
sqequalHypSubstitution,
extract_by_obid,
introduction,
cut,
Error :lambdaFormation_alt,
sqequalReflexivity,
computationStep,
sqequalTransitivity,
sqequalSubstitution
Latex:
\mforall{}k:\mBbbN{}. \mforall{}a,b,c:\mBbbZ{}.
((0 \mleq{} (a + b))
{}\mRightarrow{} ((((a * a * a) + (b * b * b)) + (c * c * c)) = k
\mLeftarrow{}{}\mRightarrow{} \mexists{}d:\mBbbN{}
(((a + b) = d)
\mwedge{} (((d = 0) \mwedge{} ((c * c * c) = k))
\mvee{} ((\mneg{}(d = 0))
\mwedge{} ((k - c * c * c rem d) = 0)
\mwedge{} (((a * a) + ((b * b) - a * b)) = ((k - c * c * c) \mdiv{} d)))))))
Date html generated:
2019_06_20-PM-02_42_04
Last ObjectModification:
2019_06_18-PM-10_47_42
Theory : num_thy_1
Home
Index