Supplement.
1. The mathematics treating a countable set isn't consistent by theorem 5.
You can prove any conjecture in it. Let's prove Riemann hypothesis(RH).
Let p1=2,p2=3,… be the prime numbers. Add only the term
P={p1}∪{p2}∪…(infinitely) to the usual language.
Set up X by this language.
Put Y=X∪{d⊂P}∪{z(a) is a meromorphic function on C.}∪
{∀x((x∈C)∧(Re x ≥1.5)→(z(x)=∏p∈d1/(1-1/px))}∪
{∀x((x∈C)∧(z(x)=0)→(Im x =0)∨(Re x =1/2))}.
Y∪{p1∈d}∪…∪{pm∈d} has a model Mm for which
Mm |= d={p1}∪…∪{pm} and is consistent for ∀m∈N. So
Y∪{p1∈d}∪{p2∈d}∪…(infinitely) is consistent and has a model M.
M is a model of X. M |= d=P={p1}∪{p2}∪…(infinitely),
M |= ∀x(z(x)=ζ(x)), M |= (RH) If X |- ¬(RH), M |= ¬(RH). This is contradiction.
You never find exceptions to (RH) by concrete calculations. This means (RH) is true.
2. For x1∋x2∋…(infinitely), {x1}∪{x2}∪…(infinitely) is countable and isn't a set.
The axiom of regularity is useless.
1. The mathematics treating a countable set isn't consistent by theorem 5.
You can prove any conjecture in it. Let's prove Riemann hypothesis(RH).
Let p1=2,p2=3,… be the prime numbers. Add only the term
P={p1}∪{p2}∪…(infinitely) to the usual language.
Set up X by this language.
Put Y=X∪{d⊂P}∪{z(a) is a meromorphic function on C.}∪
{∀x((x∈C)∧(Re x ≥1.5)→(z(x)=∏p∈d1/(1-1/px))}∪
{∀x((x∈C)∧(z(x)=0)→(Im x =0)∨(Re x =1/2))}.
Y∪{p1∈d}∪…∪{pm∈d} has a model Mm for which
Mm |= d={p1}∪…∪{pm} and is consistent for ∀m∈N. So
Y∪{p1∈d}∪{p2∈d}∪…(infinitely) is consistent and has a model M.
M is a model of X. M |= d=P={p1}∪{p2}∪…(infinitely),
M |= ∀x(z(x)=ζ(x)), M |= (RH) If X |- ¬(RH), M |= ¬(RH). This is contradiction.
You never find exceptions to (RH) by concrete calculations. This means (RH) is true.
2. For x1∋x2∋…(infinitely), {x1}∪{x2}∪…(infinitely) is countable and isn't a set.
The axiom of regularity is useless.