Metamath Proof Explorer


Theorem r19.2zb

Description: A response to the notion that the condition A =/= (/) can be removed in r19.2z . Interestingly enough, ph does not figure in the left-hand side. (Contributed by Jeff Hankins, 24-Aug-2009)

Ref Expression
Assertion r19.2zb ⊢ A ≠ ∅ ↔ ∀ x ∈ A φ → ∃ x ∈ A φ

Proof

Step Hyp Ref Expression
1 r19.2z ⊢ A ≠ ∅ ∧ ∀ x ∈ A φ → ∃ x ∈ A φ
2 1 ex ⊢ A ≠ ∅ → ∀ x ∈ A φ → ∃ x ∈ A φ
3 rzal ⊢ A = ∅ → ∀ x ∈ A φ
4 3 necon3bi ⊢ ¬ ∀ x ∈ A φ → A ≠ ∅
5 rexn0 ⊢ ∃ x ∈ A φ → A ≠ ∅
6 4 5 ja ⊢ ∀ x ∈ A φ → ∃ x ∈ A φ → A ≠ ∅
7 2 6 impbii ⊢ A ≠ ∅ ↔ ∀ x ∈ A φ → ∃ x ∈ A φ