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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 ) )