Metamath Proof Explorer


Theorem dfss5

Description: Alternate definition of subclass relationship: a class A is a subclass of another class B iff each element of A is equal to an element of B . (Contributed by AV, 13-Nov-2020)

Ref Expression
Assertion dfss5 ( 𝐴 ⊆ 𝐵 ↔ ∀ 𝑥 ∈ 𝐴 ∃ 𝑦 ∈ 𝐵 𝑥 = 𝑦 )

Proof

Step Hyp Ref Expression
1 dfss3 ⊢ ( 𝐴 ⊆ 𝐵 ↔ ∀ 𝑥 ∈ 𝐴 𝑥 ∈ 𝐵 )
2 clel5 ⊢ ( 𝑥 ∈ 𝐵 ↔ ∃ 𝑦 ∈ 𝐵 𝑥 = 𝑦 )
3 2 ralbii ⊢ ( ∀ 𝑥 ∈ 𝐴 𝑥 ∈ 𝐵 ↔ ∀ 𝑥 ∈ 𝐴 ∃ 𝑦 ∈ 𝐵 𝑥 = 𝑦 )
4 1 3 bitri ⊢ ( 𝐴 ⊆ 𝐵 ↔ ∀ 𝑥 ∈ 𝐴 ∃ 𝑦 ∈ 𝐵 𝑥 = 𝑦 )