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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Naoto Meguro. Amateur.
MSC 2010. Primary 03C62;Secondary 03C55.
Key Words and Phrases. Fermat conjecture,Beal conjecture.
The abstract. Fermat conjecture and Beal conjecture aren't proved
in the natural number theory in the formalism.

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I prove that Fermat conjecture(FC) isn't proved in the natural number theory(NNT) in the formalism.
Beal conjecture isn't proved in it too.

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Let's write ∅=0 and x∪{x}=x+1. Let X be a countable and consistent axiom system of
the mathematics like the following. Let N" be the individual symbol for which
(1∈N")∧∀x((x∈N")→(x+1∈N")) ∈X and ∀x((1∈x)∧∀y((y∈x)→(y+1∈x))→(N"⊂x)∈X.
Let d be a free individual symbol. X doesn't include d. Assume that X doesn't include the function
symbol x^y. Let ∞ be the individual symbol for which
∀x((x∈N")→(∞x=∞)) ∈X. and ∞+∞=∞ ∈X and ∞∞=∞ ∈X
Let ? be the individual symbol expressing nonsense. N"+?=?∈X etc..
And ∀x∀y((x∈N")∧(y∈N")→(x+y∈N")∧(xy∈N")∧(x+(y+1)=(x+y)+1)∧(x(y+1)=xy+x)∧(x1=x)) ∈X.
Theorem 1.X has a model M for which M |= (d∈N")∧(n≠d) for ∀n∈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 M_m |= ∞=+∞ and is consistent for ∀m∈N. So X"=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")∧(n≠d) for ∀n∈N∪{0}.♦
M |= x+1=x∪{x}∋x. Assume that M |= x∈N" and y∈N"-N for M and M |= r=x+y and M |= r"=xy.
If r∈N for M, M |= y-r-1∈N". M |= r-1=(x+y)-1=x+(y-1),…,M |= ∅=0=x+(y-r)=(x+(y-r-1))+1∋x+(y-r-1)
This is contradiction. So r∈N"-N for M.
If M |= x=1, r"=y∈N"-N for M. If M |= x≠1,r"=(x-1)y+y∈N"-N for M.
Define M" and x^y by M" |= x^y=the usual one when (M |= x∈N")∧(y∈N∪{0} for M) and
M" |= x^y=∞ when (M |= x∈N")∧(y∈N"-N for M) or (M" |= x=∞)∧(M |= y∈N") and M" |= x^y=? for the other cases
and M |= P ⇒ M"|= P. (The object domain of M" is that of M.)
If N is a set for M", M" |= N"=N but d∈N"-N for M". N isn't a set for M".
For M" and n∈N, {x|(x∈N")∧(M" |= n^x≠∞)}=N isn't a set.(You cannot write M" |= n^x≠∞ as a set-theoretical
logical formula.) You cannot prove M" |= ∀x((x∈N")→(n^x≠∞)) by induction by the axiom of subsets.
Put Y"={∀x∀y∀z((x∈N")∧(y∈N")∧(z∈N")→(x^y+z=x^yx^z)∧((x^y)^z=x^yz)∧((xy)^z=x^zy^z)),∀x((x∈N")→(x¹=x))}
and Y=X∪Y". M" is a model of X"∪Y" and Y. Y is an axiom system of NNT treating the function x^y.
Put FC=∀x∀y∀z∀m((x∈N")∧(y∈N")∧(z∈N")∧(m∈N")→(x^2+m≠y^2+m+z^2+m)).
Theorem 2. ¬(Y |- FC)
Proof. M"|= d∈N", 2+d∉N for M". M" |= 2^2+d=∞=∞+∞=2^2+d+2^2+d. M" |= ¬FC. So ¬(Y |- FC).♦
M" |= 2^d+2+3^d+3=∞+∞=∞=5^d+4. Beal conjecture isn't proved from Y.

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For M",{x|(x∈N")∧(x is a finite set.)}=N isn't a set.
You cannot write the proposition "x is a finite set." as a set-theoretical logical formula. You cannot formalize the
mathematics treating the direct sum and the tensor product. The known proof of FC uses these and isn't
formalized.

Naoto Meguro. Amateur.
MSC2010. Primary 03C62; Secondary 03C55.
Key Words and Phrases. The axiom of foundation.
The abstract. The axiom of foundation leads contradiction in the set theory.

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I indicate that the axiom of foundation(AF) leads contradiction in the standard set theory.

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Let's write ∅=0 and x∪{x}=x+1. Let X be an axiom system of the set theory.
Let N" be an individual symbol for which (1∈N")∧∀x((x∈N")→(x+1∈N")) ∈X.
This is the axiom of infinity. You may set up that ∀x(∑_m∈{x} 1=1) ∈X and
∀x∀y((x∉y)→(∑_m∈y∪{x} 1=(∑_m∈y 1)+1)) ∈X.
These are the definition of ∑_m∈x 1. You may set up that ∀x(∑_m∈N"-{x} 1=∑_m∈N" 1) ∈X
for this is formed in the standard set theory for which N"=N. (∑_m∈N-{x} 1=1+1+…(infinitely)=∑_m∈N 1)
Put s=∑_m∈N" 1.
Theorem 1. If AF∈X,X isn't consistent.
Proof. X |- s=(∑_m∈N"-{1} 1)+1=(∑_m∈N" 1)+1=s+1=s∪{s}. X |- s∈s. But X |- s∉s by AF.♦
Put s(x)=∑_m∈x 1. s(N")=s(N"-{1})∪{s(N"-{1})}∋s(N"-{1}). Similarly, s(N"-{1})∋s(N"-{1}-{2}). You get
s(N")∋s(N"-{1})∋s(N"-{1}-{2})∋s(N"-{1}-{2}-{3})∋…(infinitely). This is contradiction by AF.

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Let's remove AF. Set up that ∀x∀y((x+1=y+1)↔(x=y))∈X instead of AF.
Put ∞={0,1,2,…(infinitely){0,1,2,…{…(infinitely)…}}}={0}∪N∪{∞}.
∞∈∞ and ∞=∞+1. 0∈1∈2∈…,∈∞
Putting N"=N and s(c)=∞ for the infinite set c,you will get a model of X. X will become consistent.