My discovery about mathematics

                                  1

The substitution principle about the equality leads inconsistency of the mathematics.
Riemann hypothesis is solved(?) as a by-product. Define the propositional function P(x)
by P(t) ⇔ the term t includes a function symbol. By the substitution principle about
the equality, (ζ(-2)=0)→(P(ζ(-2))↔P(0)) is true. ζ(-2)=0 and P(ζ(-2) are true but P(0) is
false. This is contradiction.

                                  2

Let L be a language in which 0, -2, 1/2 are individual symbols and ζ(x),Re(x) are
function symbols and x=y,x≥y,P(x) are predicate symbols.
Let U={ζ(-2)=0,∀x∀y((x=y)→(P(x)↔P(y))) }∪(∪_m∈N{¬P(c_m)})∪(∪_m∈N(∪_n∈N{∀x₁…∀x_mP(f_m,n(x₁,…,x_m))}))
be an axiom system in L.
The c_m's are the individual symbols in L. The f_m,n(x₁,…x_m)'s are the function symbols in L.
Theorem 1. U isn't consistent.
Proof. U |- ζ(-2)=0. U |- (ζ(-2)=0)→(P(ζ(-2))↔P(0)). U |- P(ζ(-2)↔P(0). U |- P(ζ(-2))→P(0). U |- ∀xP(ζ(x)).
U |- P(ζ(-2)). U |- P(0). But U |- ¬P(0). U isn't consistent.♦
Let T be the standard mathematics in L. The object domain is the set of the closed terms in L.
ζ(-2)=0 is a theorem of T.
Theorem 2. T isn't consistent.
Proof. P(x) in U is a propositional function in T. The standard mathematics assumes the substitution
principle for this P(x). So ∀x∀y((x=y)→(P(x)↔P(y))) is a theorem of T with the definitions of P(x) in U.
The elements of U are theorems of T. By theorem 1, U |- P(0) and U |- ¬P(0). P(0) and ¬P(0) are theorems of T.
T isn't consistent.♦
RH=∀x((ζ(x)=0)∧(Re(x)≥0)→(Re(x)=1/2)) is a logical formula in L. RH and ¬RH are theorems of T.
RH is nonsense with the other logical formulas in L.

                                  3

The substitution principle will be wrong but you couldn't evolve the mathematics without
using it.