Metamath Proof Explorer


Theorem dfss7

Description: Alternate definition of subclass relationship. (Contributed by AV, 1-Aug-2022)

Ref Expression
Assertion dfss7 ⊢ B ⊆ A ↔ x ∈ A | x ∈ B = B

Proof

Step Hyp Ref Expression
1 dfss2 ⊢ B ⊆ A ↔ B ∩ A = B
2 dfin5 ⊢ A ∩ B = x ∈ A | x ∈ B
3 2 ineqcomi ⊢ B ∩ A = x ∈ A | x ∈ B
4 3 eqeq1i ⊢ B ∩ A = B ↔ x ∈ A | x ∈ B = B
5 1 4 bitri ⊢ B ⊆ A ↔ x ∈ A | x ∈ B = B