My discovery about mathematics

NaotoMeguro. Amateur.
MSC2010. Primary 03C62; Secondary 03C55.
Key Words and Phrases. Fermat conjecture,Beal conjecture,abc conjecture.
The abstract. Fermat conjecture and Beal conjecture and abc conjecture aren't proved
in the natural number theory in the formalism.

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Let's write ∅=0 and x∪{x}=x+1. Put p=(0∈N")∧∀x(x∈N")→(x+1∈N")) and
q=∀x((x∈N")∧(x≠0)→∃y((y∈N")∧(y+1=x))) and q"=∀x∀y((x∈N")∧(y∈N")→(x∈y)∨(x=y)∨(y∈x))
and r=∀x∀y∀z((x∈N")∧(y∈N")∧(z∈N")∧(x∈y)∧(y∈z)→(x∈z)).
Let X be a consistent and countable axiom system of the standard mathematics.
Let ∞ be the individual symbol for which 1/∞=0 ∈X and ∀x((x≥∞)→(x=∞)) ∈X
and ∞+∞=∞∈X. Let d be a free individual symbol. X doesn't include d.
Let ? be the individual symbol expressing nonsense. ∞/∞=?∈X, ∞-∞=?∈X etc..
Theorem 1. When N" is a set for which p and q and q" and r are formed,
(c∈N")∧(m∈c for ∀m∈N∪{0})⇒ 2^c=∞.
Proof. 2^c-1=∑_m∈c 2^m ≥2⁰+2¹+2²+…(infinitely)=s≥1.
If s≠∞, s-2s=1. s=-1 This is contradiction. So s=∞ s-2s=? 2^c=∞.♦
Put p"=∀x((0∈x)∧∀y((y∈x)→(y+1∈x))→(N"⊂x)).
You may set up that p,p",q,q",r ∈X.
Theorem 2. X has a model M for which (M |= d∈N")∧(M |= m∈d for ∀m∈N∪{0}.).
Proof. X∪{d∈N",0∈d,1∈d,…,m∈d} has a model M_m for which M_m |= d=m+1
and M_m |= N"=N and is consistent for ∀m∈N∪{0}.
So X∪{d∈N",0∈d,1∈d,…(infinitely)} is consistent and has a model M.
M is a model of X and (M |= d∈N")∧(M |= m∈d for ∀m∈N∪{0}.).♦
Put FC=∀x∀y∀z∀n((x∈N")∧(y∈N")∧(z∈N")∧(n∈N")→(xⁿ⁺²+yⁿ⁺²≠zⁿ⁺²)).
Theorem 3. ¬(X |- FC)
Proof. By theorem 1, M |= 2^d+2=2^d=∞ for M in theorem 2. M |= 2^d+2=∞=∞+∞=2^d+2+2^d+2.
M |= ¬FC. ¬(X |- FC)♦
M |= 2^d+2+3^d+3=∞+∞=∞=5^d+4. So Beal conjecture isn't proved from X.

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