Metamath Proof Explorer


Theorem disj3

Description: Two ways of saying that two classes are disjoint. (Contributed by NM, 19-May-1998)

Ref Expression
Assertion disj3 ⊢ A ∩ B = ∅ ↔ A = A ∖ B

Proof

Step Hyp Ref Expression
1 pm4.71 ⊢ x ∈ A → ¬ x ∈ B ↔ x ∈ A ↔ x ∈ A ∧ ¬ x ∈ B
2 eldif ⊢ x ∈ A ∖ B ↔ x ∈ A ∧ ¬ x ∈ B
3 2 bibi2i ⊢ x ∈ A ↔ x ∈ A ∖ B ↔ x ∈ A ↔ x ∈ A ∧ ¬ x ∈ B
4 1 3 bitr4i ⊢ x ∈ A → ¬ x ∈ B ↔ x ∈ A ↔ x ∈ A ∖ B
5 4 albii ⊢ ∀ x x ∈ A → ¬ x ∈ B ↔ ∀ x x ∈ A ↔ x ∈ A ∖ B
6 disj1 ⊢ A ∩ B = ∅ ↔ ∀ x x ∈ A → ¬ x ∈ B
7 dfcleq ⊢ A = A ∖ B ↔ ∀ x x ∈ A ↔ x ∈ A ∖ B
8 5 6 7 3bitr4i ⊢ A ∩ B = ∅ ↔ A = A ∖ B