Metamath Proof Explorer


Theorem ssel

Description: Membership relationships follow from a subclass relationship. (Contributed by NM, 5-Aug-1993) Avoid ax-12 . (Revised by SN, 27-May-2024)

Ref Expression
Assertion ssel ( 𝐴 ⊆ 𝐵 → ( 𝐶 ∈ 𝐴 → 𝐶 ∈ 𝐵 ) )

Proof

Step Hyp Ref Expression
1 df-ss ⊢ ( 𝐴 ⊆ 𝐵 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) )
2 id ⊢ ( ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) → ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) )
3 2 anim2d ⊢ ( ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) → ( ( 𝑥 = 𝐶 ∧ 𝑥 ∈ 𝐴 ) → ( 𝑥 = 𝐶 ∧ 𝑥 ∈ 𝐵 ) ) )
4 3 aleximi ⊢ ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) → ( ∃ 𝑥 ( 𝑥 = 𝐶 ∧ 𝑥 ∈ 𝐴 ) → ∃ 𝑥 ( 𝑥 = 𝐶 ∧ 𝑥 ∈ 𝐵 ) ) )
5 dfclel ⊢ ( 𝐶 ∈ 𝐴 ↔ ∃ 𝑥 ( 𝑥 = 𝐶 ∧ 𝑥 ∈ 𝐴 ) )
6 dfclel ⊢ ( 𝐶 ∈ 𝐵 ↔ ∃ 𝑥 ( 𝑥 = 𝐶 ∧ 𝑥 ∈ 𝐵 ) )
7 4 5 6 3imtr4g ⊢ ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) → ( 𝐶 ∈ 𝐴 → 𝐶 ∈ 𝐵 ) )
8 1 7 sylbi ⊢ ( 𝐴 ⊆ 𝐵 → ( 𝐶 ∈ 𝐴 → 𝐶 ∈ 𝐵 ) )