return false

← 2 messages moved from Basic Mathematics

hey :)

@VyxalBot hello

I know you won't respond because you're busy serving us in Vyxal chat, but I wanted to say hi anyway ;)

Ou, I can play here O.O, this is so much fun

> it would seem that V[ω[1]] (see below) is a model for any statement that can (fairly) be said to have been implied by the rough description, and not all of the axioms of replacement hold in V[ω[1]]. Worse yet, Vδ1 would also seem to be such a model. (δ1 is the first uncomputable ordinal.) Thus the axioms of replacement do not seem to us to follow from the iterative conception.

← 1 message moved from Logic

∀ x,y,z ∈ ℝ ( x < y ∧ z < 0 ⇒ x·z > y·z ) [lemma]
  Given x,y ∈ ℝ:
    If x < y ∧ z < 0:
      -z > 0
      x·(-z) > y·(-z)
      x·(-z) +(-x·(-z))+(-y(-z)) < -y·(-z)+y·(-z)+(-x·(-z))
      -y·(-z) < -x·(-z)
      -x·(-z) > -y·(-z)
      x·z > y·z
  ∀ x,y,z ∈ ℝ ( x < y ∧ z < 0 ⇒ x·z > y·z )

There is an error.

Feel free to delete. Now, this is my attempt.

← 3 messages moved from Basic Mathematics