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F. Zer

01:43

Prove ∀ S ∈ set ( S ≠ ∅ ⇔ ∃ x ∈ obj ( x ∈ S ) ) [Lemma]
  Given S ∈ set:
    ∀ S,T ∈ set ( S ≠ T ⇔ ¬∀x ∈ obj ( x ∈ S ⇔ x ∈ T ) ) [Lemma]
      Given S,T ∈ set:
        If S ≠ T:
          If ∀x ∈ obj ( x ∈ S ⇔ x ∈ T ):
            ∀ S,T ∈ set ( S = T ⇔ ∀x ∈ obj ( x ∈ S ⇔ x ∈ T ) )
            S = T
            ⊥
          ¬∀x ∈ obj ( x ∈ S ⇔ x ∈ T )
        If ¬∀x ∈ obj ( x ∈ S ⇔ x ∈ T ):
          If S = T:
            ∀ S,T ∈ set ( S = T ⇔ ∀x ∈ obj ( x ∈ S ⇔ x ∈ T ) )
            ∀x ∈ obj ( x ∈ S ⇔ x ∈ T )

Please, delete the post above (and this one).

user

02:12

@lyxal not sure what that means but ok

user21820

03:03

← 2 messages moved from Basic Mathematics

10 hours later…

F. Zer

13:11

Given S ∈ set:
If S ≠ ∅:
∀ S ∈ set ( S ≠ ∅ ⇔ ∃ x ∈ obj ( x ∈ S ) ) [Lemma]
∃ x ∈ obj ( x ∈ S )
Let A ∈ obj such that A ∈ S
Let I' = { x : x ∈ A }
Given x ∈ obj:
If x ∈ I':
Given T ∈ S:

x ∈ T
∀x∈obj ( x∈I' ⇔ ∀T∈S ( x∈T ) )
∃I∈set ∀x∈obj ( x∈I ⇔ ∀T∈S ( x∈T ) )

2 hours later…

user21820

15:02

← 1 message moved from Basic Mathematics

3 hours later…