Metamath Proof Explorer


Theorem disjeq

Description: Equality theorem for disjoints. (Contributed by Peter Mazsa, 22-Sep-2021)

Ref Expression
Assertion disjeq ( 𝐴 = 𝐵 → ( Disj 𝐴 ↔ Disj 𝐵 ) )

Proof

Step Hyp Ref Expression
1 eqimss2 ⊢ ( 𝐴 = 𝐵 → 𝐵 ⊆ 𝐴 )
2 1 disjssd ⊢ ( 𝐴 = 𝐵 → ( Disj 𝐴 → Disj 𝐵 ) )
3 eqimss ⊢ ( 𝐴 = 𝐵 → 𝐴 ⊆ 𝐵 )
4 3 disjssd ⊢ ( 𝐴 = 𝐵 → ( Disj 𝐵 → Disj 𝐴 ) )
5 2 4 impbid ⊢ ( 𝐴 = 𝐵 → ( Disj 𝐴 ↔ Disj 𝐵 ) )