My discovery about mathematics

                             1

Let's see that the model theory proves inconsistency of the mathematics.
The substitution principle about the equality is wrong.

                             2

Put A=Z/2Z={0,1}. Let L be the language in which the individual symbols are only
0 and 1 and the function symbol is only f(x,y)=x+y and x=y is a predicate symbol..
Assume that theory of the ring A is consistent. Let X={0≠1} be an axiom system in L.
Let M be a model of X in which the object domain is the ring A, M |= 1=1, M |= 0=0, M |=¬(0=1),
M |= ¬(1=0) , and M |= ∀x∀y(f(x,y)=0xy=0).
Let T be the theory in which the terms are terms in L.
P(x)⇔ (M |= x=0) is a propositional function in T. P(0), ¬P(1), ∀x∀y(P(x+y)) are true in T.
So ∀x∀y((x=y)→(P(x)↔P(y))) is true in T. (The mathematics needs this axiom about =
though this is doubtful as you will see.)
Theorem 1. 1+0=1 is false in T.
Proof. (1+0=1)→(P(1+0)↔P(1)) is true in T. P(1) is false. P(1+0) is true. P(1+0)↔P(1)
is false. So 1+0=1 is false in T.♦
Let T" be the theory of the module A in L.
Theorem 2. T" isn't consistent.
Proof. The terms in T" are terms in L. By theorem 1, 1+0≠1 in T".
1+0=1 in T" for A is a module in L. This is contradiction.♦
Another proof of theorem 2. Let Q(x) be a predicate symbol in L which represents
the propositional function P(x). Q(0) and ¬Q(1) and ∀x∀yQ(x+y) are true in T".
If T" is consistent, T" is a model of U={ 1+0=1, (1+0=1)→(Q(1+0)↔Q(1)),
Q(0), ¬Q(1), ∀x∀yQ(x+y) }. U is consistent then. U |- 1+0=1,
U |- (1+0=1)→(Q(1+0)↔Q(1)), U |- Q(1+0)↔Q(1). U |- Q(1+0)→Q(1).
U |- Q(1+0), U |- Q(1). But U |- ¬Q(1), U isn't consistent. This is contradiction.♦
The theory of the ring A includes the theory of the module A and isn't consistent.
The mathematics treating the ring A isn't consistent and proves every conjecture
and the denial of it. The millennium problems and abc conjecture are proved by
using the ring A.

                             3

The above may be ridiculous but isn't a fake. The mathematics which you know is
nonsense for the time being.

                             1

Let's see inconsistency of the mathematics.

                             2

Let X={0=0,1=1,0≠1,1≠0, M(0,0),M(1,1), ¬M(1,0),¬M(0,1),M(z(1),0)}
be an axiom system (z(x) is an arbitrary function symbol in the mathematics.)
∀x∀y((x=y)→(P(x)↔P(y))) ( P(x) is a logical formula in the mathematics.) is an axiom
of the mathematics.
Theorem 1.X is consistent.
Proof. Define a model of X by M |= z(1)=0 and M |= M(x,y)⇔ x=y etc..♦
The mathematics includes the theory by X.
Theorem 2. z(1)≠1 in the mathematics.
Proof. ∀x∀y((x=y)→(M(x,0)↔M(y,0))) is true in the mathematics.
(z(1)=1)→(M(z(1),0)↔M(1,0)). (z(1)=1)→(M(z(1),0)→M(1,0)).
M(z(1),0)∧(¬M(1,0))→(z(1)≠1). The mathematics includes M(z(1),0) and ¬M(1,0) as axioms.
z(1)≠1 is proved in the mathematics.♦
Theorem 3. The mathematical theory in which 1+0=1 isn't consistent
Proof. By theorem 2.(Put z(x)=x+0. z(x) was arbitrary.)♦
The theory of the complex function isn't consistent and proves Riemann hypothesis
and BSD conjecture and the denial of them.♦

                            3

The above may be ridiculous but isn't a fake.