My discovery about mathematics

Let c={c₁,c₂…} be a countable set.
Add the propositional function p(x)=(x=c₁)∨(x=c₂)∨…(infinitely) to the usual language.
The closed logical formulas are still countable. You can use Gödel's theorem.
Let X be an axiom system by the language.
Let d be a free individual symbol. The mathematical axioms in X don't include d.
Put X" =X ∪{∀xp(x)}.
Theorem 1. X" isn't consistent.
Proof. If X" is consistent, X"∪{d≠c₁}∪…∪{d≠c_m} has a model M_m for which
M_m |= d=c_m+1 and is consistent for ∀m∈N.
So X"∪{d≠c₁}∪{d≠c₂}∪…(infinitely) is consistent and has a mdel M by Göel's theorem.
M |= d≠c₁, M |= d≠c₂,…(infinitely)
The object domain of M is larger than c. But M |= ∀xp(x). This is contradiction.♦

Let c={c₁,c₂,…} (c_i≠c_k for i≠k) be a countable set.
Add the propositional functions p_m(x)=(x=c_m)∨(x=c_m+1)∨…(infinitely) (m∈N)
and q(x)=p₂(x)∧p₃∧…(infinitely) to the usual language.
The closed logical formulas are still countable. You can use Gödel's theorem.
Let X be a countable axiom system by the language.
Let d be a free individual symbol. The mathe matical axioms in X don't include d.
Theorem1. X isn't consistent.
Proof. p_m+1(c_m) is false. So ¬q(c_m) is true.
If X is consistent, X∪{p₁(d)}∪{¬q(d)}∪{d≠c₁}∪…∪{d≠c_m} has
a model M_m for which M_m |= d=c_m+1 and is consistent for ∀m∈N.
So X∪{p₁(d)}∪{¬q(d)}∪{d≠c₁}∪{d≠c₂}∪…(infinitely) is consistent and has
a model M by Gödel's theorem. p_m(x)=p₂(x)∧…∧p_m(x)
So M |= p₁(d) ⇔ (M |= d=c₁)∨(M |= p₂) ⇔ (M |= d=c₁)∨(M |= d=c₂)∨(M |= p₂(d)∧p₃(d))⇔…
Repeating this infinitely,you get (M |= d=c₁)∨(M |= d=c₂)∨…(infinitely)∨(M |= q(d)).
But M |= d≠c₁, M |= d≠c₂,…(infinitely),M |= ¬q(d). This is contradiction.♦
Theorem 2. You can trump up a proof of RH.
Proof. If the c_m's are the zero points of ζ(x), RH=∀x(p₁(x)→(Re x=1/2)∨(Im x=0)).
You are to use X.
X |- RH and X |- ¬RH by theorem 1.
Theorem 3. You can trump up proofs