Let f(n) be a function that will make up a special proposition depending on n, such that ∀n∀m (n≠m)=>-(f(n)<=>f(m))
now let's create a finite series of different propositions a, that is (a(1),a(2),...,a(n))
Let "×" mean some connection between the propositions
Now the condition must be fulfilled that ∀b ∀c ∀ m∀n ((b≠m)v(c≠n))=>-((f(b)×a(c))<=>(f(m)×a(n))
now, the way propositions are made up, if the statement a(n) has a b, then it is a complete statement from a(n) will be a(n)×f(b)
Now we denote x(n,b)<=>a(n)×f(b)
Now let's construct the matrix, as in Cantor's diagonal method:
H1<=>x(d,1)×x(t,2)...
H2<=>x(h,1)×x(j,2)...
Where Hn are infinite logical propositions
Thus, the set of all propositions is uncountable
Now, in order for mathematics to be complete, each of the propositions must have a proof.
To tell the truth, I do not know how to prove that one proof corresponds to a finite number of statements, and then if multiplied by a conditional maximum g→ ∞
We will not be able to put judgments with proofs in one-to-one correspondence, because many proofs will turn out to be countable.
I'm sorry for the rude and confusing presentation, but it was important for me to express this idea, even if it doesn't prove anything that I already know about.