Metamath Proof Explorer


Theorem ssrd

Description: Deduction based on subclass definition. (Contributed by Thierry Arnoux, 8-Mar-2017)

Ref Expression
Hypotheses ssrd.0 ⊢ Ⅎ 𝑥 𝜑
ssrd.1 ⊢ Ⅎ 𝑥 𝐴
ssrd.2 ⊢ Ⅎ 𝑥 𝐵
ssrd.3 ⊢ ( 𝜑 → ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) )
Assertion ssrd ( 𝜑 → 𝐴 ⊆ 𝐵 )

Proof

Step Hyp Ref Expression
1 ssrd.0 ⊢ Ⅎ 𝑥 𝜑
2 ssrd.1 ⊢ Ⅎ 𝑥 𝐴
3 ssrd.2 ⊢ Ⅎ 𝑥 𝐵
4 ssrd.3 ⊢ ( 𝜑 → ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) )
5 1 4 alrimi ⊢ ( 𝜑 → ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) )
6 2 3 dfssf ⊢ ( 𝐴 ⊆ 𝐵 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) )
7 5 6 sylibr ⊢ ( 𝜑 → 𝐴 ⊆ 𝐵 )