Metamath Proof Explorer


Theorem disj

Description: Two ways of saying that two classes are disjoint (have no members in common). (Contributed by NM, 17-Feb-2004) Avoid ax-10 , ax-11 , ax-12 . (Revised by GG, 28-Jun-2024)

Ref Expression
Assertion disj ( ( 𝐴𝐵 ) = ∅ ↔ ∀ 𝑥𝐴 ¬ 𝑥𝐵 )

Proof

Step Hyp Ref Expression
1 df-in ( 𝐴𝐵 ) = { 𝑦 ∣ ( 𝑦𝐴𝑦𝐵 ) }
2 1 eqeq1i ( ( 𝐴𝐵 ) = ∅ ↔ { 𝑦 ∣ ( 𝑦𝐴𝑦𝐵 ) } = ∅ )
3 eleq1w ( 𝑦 = 𝑥 → ( 𝑦𝐴𝑥𝐴 ) )
4 eleq1w ( 𝑦 = 𝑥 → ( 𝑦𝐵𝑥𝐵 ) )
5 3 4 anbi12d ( 𝑦 = 𝑥 → ( ( 𝑦𝐴𝑦𝐵 ) ↔ ( 𝑥𝐴𝑥𝐵 ) ) )
6 5 eqabcbw ( { 𝑦 ∣ ( 𝑦𝐴𝑦𝐵 ) } = ∅ ↔ ∀ 𝑥 ( ( 𝑥𝐴𝑥𝐵 ) ↔ 𝑥 ∈ ∅ ) )
7 imnan ( ( 𝑥𝐴 → ¬ 𝑥𝐵 ) ↔ ¬ ( 𝑥𝐴𝑥𝐵 ) )
8 noel ¬ 𝑥 ∈ ∅
9 8 nbn ( ¬ ( 𝑥𝐴𝑥𝐵 ) ↔ ( ( 𝑥𝐴𝑥𝐵 ) ↔ 𝑥 ∈ ∅ ) )
10 7 9 bitr2i ( ( ( 𝑥𝐴𝑥𝐵 ) ↔ 𝑥 ∈ ∅ ) ↔ ( 𝑥𝐴 → ¬ 𝑥𝐵 ) )
11 10 albii ( ∀ 𝑥 ( ( 𝑥𝐴𝑥𝐵 ) ↔ 𝑥 ∈ ∅ ) ↔ ∀ 𝑥 ( 𝑥𝐴 → ¬ 𝑥𝐵 ) )
12 2 6 11 3bitri ( ( 𝐴𝐵 ) = ∅ ↔ ∀ 𝑥 ( 𝑥𝐴 → ¬ 𝑥𝐵 ) )
13 df-ral ( ∀ 𝑥𝐴 ¬ 𝑥𝐵 ↔ ∀ 𝑥 ( 𝑥𝐴 → ¬ 𝑥𝐵 ) )
14 12 13 bitr4i ( ( 𝐴𝐵 ) = ∅ ↔ ∀ 𝑥𝐴 ¬ 𝑥𝐵 )