My discovery about mathematics

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Let's see defect of the mathematics treating two elements.

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Let X={0≠1} be an axiom system. Set up that the predicate symbol is only = and the
function symbol is only z(x). Assume that X is consistent. Define a model of X whose
object domain is {0,1} by ∀s(z(s)=0). Let M be it.
Put P=∀s(z(s)=s), Q=∀s((z(s)=0)→(s=0)).
Theorem 1. M |= Q is false.
Proof. M |= z(1)=0 is true. M |= 1≠0. M |= ¬Q is true.♦
Let L be the logic treating "=" and the model of X whose object domain is {0,1}.
Theorem 2. P is false in L.
Proof. ∀s∀t((t=s)→((M |= (t=0)→(s=0))↔(M |= (s=0)→(s=0)))) is an axiom of L.
(1=1)→((M |= (1=0)→(1=0))↔(M |= (1=0)→(1=0)))) is true.
If (z(1)=1, (z(1)=1)→((M |= (z(1)=0)→(1=0))↔((M |= (1=0)→(1=0)))).
If z(0)=0, (z(0)=0)→((M |= (z(0)=0→(0=0))↔((M |= (0=0)→(0-0)))).
is true too.
∀s((z(s)=s)→((M |= (z(s)=0)→(s=0))↔(M |= (s=0)→(s=0)))) is true..
∀s(z(s)=s)→∀s((M |= (z(s)=0)→(s=0))↔(M |= (s=0)→(s=0)))).
∀s(z(s)=s)→(∀s(M |= (s=0)→(s=0))→∀s(M |= (z(s)=0)→(s=0))).
P→((M |= ∀s((s=0)→(s=0)))→(M |= Q)). M |= Q is false by theorem 1.
M |= ∀s((s=0)→(s=0)) is true. So P is always false in L. ♦
The models of X whose object domain is {0,1} are decided by z(0) and z(1).
Set up that z(0)=0 and z(1)=1. It makes a model of X whose object domain is
{0,1}. L treats it. z(0)=0 and z(1)=1 in L then. This is contradiction by theorem 2.
X or L isn't consistent. The theory of Z/2Z is a model of X whose
object domain is {0,1}. L treats it. Define z(x)=1x. ∀s(1x=x) is false by theorem 2.
The theory of the commutative algebra like Z/2Z isn't consistent.

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..
The above may be ridiculous but isn't a fake. Someone must overcome this
paradox.