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Goldbach's Conjec u e — Towa ds he Inconsis ency o A i hme ic
Ral Wüs ho en
Abs ac . This pape p o es ha ZFC and Peano a i hme ic (PA) a e inconsis en , he la e
esul being a co olla y o he o me . We in oduce a me a heo y o e ZFC ha allows us o
use s a emen s in he p oo ha exp ess logical consequence. We hen show, by explici ly
s a ing a con adic ion, ha his me a heo y is inconsis en , which immedia ely leads o he
inconsis ency o ZFC. The con adic ion i sel is igge ed by he conjunc ion o wo p ope ies
o an in ini e se , wi h which we exp ess a s eng hened o m o he s ong Goldbach
conjec u e. We use elemen a y numbe heo y, whe e he cons uc i e ole o p ime numbe s
wi hin he na u al numbe s is a key poin .
No a ions. Le deno e he na u al numbe s s a ing om 1, le a deno e he na u al
numbe s s a ing om a > 1 and le 3 deno e he p ime numbe s s a ing om 3.
Le ZFCμ be he me a heo y o e ZFC ha is de ined as he s anda d me a heo y o e ZFC
(ZFC wi h Gödel-numbe ed syn ax and de inable p o abili y p edica e), whe e ins ead o he
syn ac ic en ailmen ⊢ P (i.e., " he e exis s a p oo o P"), we use he s a emen
"we ha e a p oo o P", which means ha his pape con ains a p oo in ZFC ha P holds. We
symbolize "we ha e a p oo o P" by ⊢w P.
Le SSGB deno e he ollowing s eng hened o m o he s ong Goldbach conjec u e: E e y
e en numbe g ea e han 6 is he sum o wo dis inc odd p imes.
Theo em. ZFC is inconsis en . [1], [2]
P oo . In he ollowing we use ZFCμ and p oduce a o mula φ along wi h bo h a p oo o φ
om ZFCμ and a p oo o φ om ZFCμ. We hen show ha he ZFCμ inconsis ency
immedia ely implies he inconsis ency o ZFC.
We de ine he se Sg := { (pk, mk, qk) | k, m ; p, q 3, p < q; m = (p + q) / 2 }.
Sg has he ollowing wo p ope ies.
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Fi s , he whole ange o 3 can be exp essed by he iple componen s o Sg (”co e ing”).
We p o e his by di iding i in o he ollowing h ee cases.
(i) x 3 is p ime. Then, x = pk wi h p 3, k = 1.
(ii) x 3 is composi e and no a powe o 2. Then, x = pk wi h p 3, k ≠ 1.
(iii) x 3 is a powe o 2. Then, x = (p + q)k / 2 wi h p = 3, q = 5, k = (a powe o 2).
So we ha e
(C) x 3 Ǝ (pk, mk, qk) Sg x = pk x = mk.
Second, all pai s (p, q) o dis inc odd p imes a e used in he de ini ion o he se Sg
(“maximali y”). So we ha e
(M) p, q 3, p < q k (pk, mk, qk) Sg, whe e m = (p + q) / 2.
SSGB is equi alen o saying ha e e y in ege g ea e han o equal o 4 is he a i hme ic
mean o wo dis inc odd p imes. So, unde he assump ion SSGB he e is an n 4 ha
is di e en om all m de ined in Sg, whe eas unde he assump ion SSGB he e is no such n.
The ollowing s eps a e independen o he choice o n i he e is mo e han one. Fo example,
he minimal such n wo ks.
The p ope y (C) implies ha o he abo e n, e e y nk', k' , equals a componen o some
Sg iple.
The p ope y (M) excludes he possibili y ha n is he a i hme ic mean o a pai o dis inc odd
p imes no used in Sg. So, (M) excludes he possibili y ha he ques ion o whe he SSGB
holds o no depends on whe he (M) holds o no .
The basic idea is now he ollowing.
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Due o he p ope ies (C) and (M), om he assump ion SSGB i logically ollows ha he
se Sg is he union o he ollowing iples.
(a) Sg iples o he o m (pk = nk', mk, qk) wi h k = k', i n is p ime
(b) Sg iples o he o m (pk = nk', mk, qk) wi h k ≠ k', i n is composi e and no a powe o 2
(c) Sg iples o he o m (3k, 4k = nk', 5k), i n is a powe o 2
(d) all o he Sg iples o he o m (pk = nk', mk, qk), (pk, mk = nk', qk) o (pk, mk, qk = nk')
(e) Sg iples o he o m (pk ≠ nk', mk ≠ nk', qk ≠ nk').
F om he assump ion SSGB i logically ollows ha he se Sg is he abo e union o iples,
whe e n is eplaced by any y 3.
The e o e, he Sg iples a e he same in he case whe e we assume ha n exis s (i.e.,
SSGB) and in he case whe e we assume ha n does no exis (i.e., SSGB). This is
con adic ed by he ac ha unde he assump ion SSGB he numbe s m de ined in Sg ake
all in ege alues ≥ 4 whe eas unde he assump ion SSGB hey don’ .
To o malize his idea, we p oceed as ollows.
We spli Sg in o wo complemen a y subse s in he ollowing way. Fo any y 3, we w i e
Sg = Sg+(y) ∪ Sg-(y), wi h
Sg+(y) := { (pk, mk, qk) Sg | Ǝ k' pk = yk' mk = yk' qk = yk' }
Sg-(y) := { (pk, mk, qk) Sg | k' pk ≠ yk' mk ≠ yk' qk ≠ yk' }.
We de ine S1 := { (pk, mk, qk) Sg | SSGB } and S2 := { (pk, mk, qk) Sg | SSGB }. I.e.,
S1 = Sg i SSGB is ue, and S1 = { } i SSGB is alse
and
S2 = Sg i SSGB is ue, and S2 = { } i SSGB is alse.
Then, since unde bo h assump ions SSGB and SSGB he p ope ies (C) and (M) hold, we
ob ain
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(1.1) ⊢w y 3 ( SSGB => S1 = Sg+(y) ∪ Sg-(y) )
(1.2) ⊢w ( SSGB => S2 = Sg+(n) ∪ Sg-(n) ).
So, since Sg+(n) ∪ Sg-(n) is independen o n,
(1.1') ⊢w y 3 ( SSGB => S1 = Sg+(y) ∪ Sg-(y) )
(1.2') ⊢w y 3 ( SSGB => S2 = Sg+(y) ∪ Sg-(y) ).
Now, we make use o he ollowing p inciple.
I wo se s o (possibly in ini ely many) z- uples a e equal, hen he se s o hei co esponding
i- h componen s a e equal; 1 ≤ i ≤ z.
Fo his we de ine M1 := { m | (p, m, q) S1 } and M2 := { m | (p, m, q) S2 }.
Then, applying he p inciple abo e o he middle componen o he iples (p, m, q), he ac
ha we ha e a p oo o each o he implica ions in ( (1.1') (1.2') ) implies by ansi i i y
(2.1) ⊢w y 3 ( SSGB => M1 = { m | (p, m, q) Sg+(y) ∪ Sg-(y) } )
(2.2) ⊢w y 3 ( SSGB => M2 = { m | (p, m, q) Sg+(y) ∪ Sg-(y) } ).
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Now we make use o he ollowing me ama hema ical analogue o he i s -o de logic ule
x ( P(x) Q(x) ) <=> ( x P(x) ) ( x Q(x) ).
Lemma. (Dis ibu ion o Uni e sal Quan i ie o e Conjunc ion in P oo s)
Le T be a o mal heo y, and le P(x) and Q(x) be o mulas. Then,
T ⊢ x ( P(x) Q(x) ) <=> ( T ⊢ x P(x) ) ( T ⊢ x Q(x) ).
(The lemma holds in s anda d p oo sys ems o i s -o de logic; see s anda d ex books, e.g.
[3]).
Applying he lemma o ZFC and ⊢w, we ob ain ha ( (2.1) (2.2) ) is equi alen o
(2) ⊢w y 3
( ( SSGB => M1 = { m | (p, m, q) Sg+(y) ∪ Sg-(y) } )
( SSGB => M2 = { m | (p, m, q) Sg+(y) ∪ Sg-(y) } ) ).
We de ine M := { m | (p, m, q) Sg }. Then, since o e e y y 3 Sg+(y) ∪ Sg-(y) equals Sg
by de ini ion, o e e y y 3 { m | (p, m, q) Sg+(y) ∪ Sg-(y) } equals M by de ini ion.
I SSGB is ue, M is equal o 4, and i SSGB is alse, M is equal o some non-emp y p ope
subse U o 4.
I ollows ha he e is exac ly one se X { 4, U } ha { m | (p, m, q) Sg+(y) ∪ Sg-(y) } is
equal o. Since o e e y y 3 { m | (p, m, q) Sg+(y) ∪ Sg-(y) } = X ega dless o whe he
SSGB o SSGB holds, in (2) we can eplace { m | (p, m, q) Sg+(y) ∪ Sg-(y) } by X.
Then, since, analogous o ⊢, ⊢w dis ibu es o e conjunc ion, we ob ain
(3) Ǝ! X { 4, U } ( ⊢w ( SSGB => M1 = X ) ⊢w ( SSGB => M2 = X ) ).
Since he s a emen s ( SSGB => M1 = X ) and ( SSGB => M2 = X ) depend on X and
since X is he unique elemen o { 4, U } such ha hese s a emen s hold, we will make use
o he ollowing ule.
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Le P1(A) and P2(A) be s a emen s ha depend on a se A. Le A be he unique elemen o
{ B1, B2, ..., Bz } such ha P1(A) and P2(A) hold.
Then,
( Ǝ! A { B1, B2, ..., Bz } ( ⊢w P1(A) ⊢w P2(A) ) ) => ( ( ⊢w P1(B1) ⊢w P2(B1) )
( ⊢w P1(B2) ⊢w P2(B2) ) ... ( ⊢w P1(Bz) ⊢w P2(Bz) ) ).
We apply he abo e ule wi h
P1(A) = ( SSGB => M1 = A ), P2(A) = ( SSGB => M2 = A )
z = 2, B1 = 4, B2 = U.
Then, since he le -hand side o he ule is ue, we ob ain
(3.1) ( ⊢w ( SSGB => M1 = 4 ) ⊢w ( SSGB => M2 = 4 ) )
(3.2) ( ⊢w ( SSGB => M1 = U ) ⊢w ( SSGB => M2 = U ) ).
This implies
(4.1) ⊢w ( SSGB => M2 = 4 )
(4.2) ⊢w ( SSGB => M1 = U ).
Now, we will es ablish a con adic ion o ( (4.1) (4.2) ).
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We ha e a p oo o ( SSGB => M = 4 ) and we ha e a p oo o ( SSGB => M = U ≠ 4 ).
The e o e, since SSGB => M1 = M and SSGB => M2 = M by de ini ion, we ge
(5.1) ⊢w ( SSGB => M1 = 4 )
(5.2) ⊢w ( SSGB => M2 = U ≠ 4 ).
Because o ( (5.1) (5.2) ) and because
⊢w ( SSGB => M2 = { } ≠ 4 )
and
⊢w ( SSGB => M1 = { } ≠ U ),
we ha e a p oo ha ( M2 = 4 ) is alse and we ha e a p oo ha ( M1 = U ) is alse.
The e o e, ( (4.1) (4.2) ) yields
(6.1) ⊢w ( SSGB => FALSE )
(6.2) ⊢w ( SSGB => FALSE ).
And his yields
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(7.1) ⊢w SSGB
(7.2) ⊢w SSGB.
On he o he hand, since his pape con ains nei he a p oo o SSGB no o SSGB, he
nega ion o ( (7.1) (7.2) ) also holds. This p o es ha ZFCμ is inconsis en .
Since de ini ional ex ensions o ZFC p ese e consis ency, we ha e ha Con(ZFC) =>
Con(ZFCμ). By con aposi ion we ob ain ha ZFC is inconsis en .
□
Co olla y. Peano a i hme ic (PA) is inconsis en .
P oo . The e m Sg om he abo e inconsis ency p oo is no a s anda d pa o PA, bu i
can easily be de ined wi hin PA. This also applies o all o he se s used in ha p oo , since
hey a e all based on Sg o on . The e o e, he co olla y is p o ed in he same way by using
he analogous ex ension PAμ o PA.
□
Re e ences
[1] Wa ning Signs o a Possible Collapse o Con empo a y Ma hema ics, by Edwa d Nelson
(2006). h ps://web.ma h.p ince on.edu/~nelson/pape s/wa n.pd
[2] The Consis ency o A i hme ic, by Timo hy Y. Chow (2018). h ps://a xi .o g/pd /1807.05641
[3] In oduc ion o Me ama hema ics, by S ephen Cole Kleene (No h-Holland, 1952).
