User:Gistya/sandbox

The proof of the Riemann Hypothesis is as follows: Let H be a non-negative integer, and let H’ be a single-parameter family of functions on H that satisfy the following conditions:

(1) $$ \text{∀}x(\text{H}x)(\exists y(\text{H}y)(\exists z(\text{H}z)(\exists xz(\text{H}x))(\exists yz(\text{H}y)(\exists zz(\text{H}z)))\text{⊢}\text{H}xyz(\exists yzz(\text{H}y)). $$

(2) $$ {\displaystyle \forall x(\mathbf {H} _{x})(\simeq \mathbf {H} _{y})(\simeq \mathbf {Z} _{z})(\simeq xz(H_{x})(\simeq \mathbf {H} _{y})(\simeq zz(H_{y}))).} $$

By definition, a function that satisfies (1) and (2) must have a minimum value (that is, its functional graph must have a vertex in which all sides are less than or equal to zero), since it would not be a function at all if it had a non-zero minimum value. Thus a function that satisfies (1) and (2) must be bounded and/or monotonic. For this proof, we ignore this step and focus on the further proof that (2).

Clearly if H’ has a minimum value then so must H. The minimum value H must be defined for all x ∈ ℝ. This can be done with a finite or an infinite Lebesgue measure, or by using the Lebesgue integral, or by using Riemann–Liouville theory.

In a finite set, the minimum value is the point with the smallest Lebesgue measure. The minimal Lebesgue measure on a finite set is denoted with a bar through it. The resulting measure measures the minimum value, as well as the value at some other minimum; hence the name minimal measure.

When H is non-empty, it is possible to evaluate H using the Lebesgue measure, giving the following argument. Let H be the set of all functions on H such that for any two functions h ( x ) and h’ ( x ) and any monotone function f ( x ) the Lebesgue measure of f is (h x − h h’ x ) for some function h’ ( x ) and monotone function f. Now h’ ( x ) = h x − h f ( x ) for any monotone function f and any such that f(x) > f(x) for all x > 0.

As before, we find that the Lebesgue measure of H is the minimum Lebesgue measure on H. But now, suppose the domain of H is ℝ, and suppose H’ is the Lebesgue measure defined for all non-empty subsets of ℝ. By definition, this Lebesgue measure defines a unique Lebesgue measure 0 that is equal to the Lebesgue measure on H. By letting H be the non-empty subset of ℝ, we conclude that H’ has the minimum Lebesgue measure, and hence that H is bounded and non-empty.

Informal proof
When the Lebesgue measure of H is 0, H has a smallest positive integral. If H is non-empty then H’ must be of positive measure, and hence there exists a positive measure 0 with this property, which is denoted ∞.

The total amount of space occupied by the closed interval (0,∞) cannot be greater than the Lebesgue measure of H. Hence ∞ = 1 for all subsets of H. Hence ∞ = ∞.

But for some subsets of H, H H’ = ∞ for some measurable function

g ( x ) ∈ H g ′ ( x ) ∈ H

and for any real number $x \text{∈} H$

where g ′ ( x ) = ∑ y ∈ ℝ g ( x − y ) − g ( x ) g ′ ( x ) g ( x ), i.e. the Lebesgue measure of g ( x ) for all subsets of H H’ = ∞.

A formula is then

$$ h(x)=h\infty \infty (1-H \infty \circ \infty ) $$

$$ {\displaystyle h(x)=h^{{\infty }^{\infty }}\left(1-H^{{\infty }^{{\circ }^{\circ }}}\right).} $$

This formula can be found in the paper of Gödel, Peano and Hilbert. It has no intuitive meaning, but it is, by Gödel’s incompleteness theorems, a strict valid proof of the consistency of the consistency lemma for Arrow’s axiom, for any version of the consistency lemma. More formally, in the proof given above the single largest positive real root of $h(x)$ was (as had been shown by Tarski) some open interval whose Lebesgue measure was defined by the absolute values of the Lebesgue measures of $x 0$  and $y 0$.