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Metamath Proof Explorer

Theorem rabssf

Description: Restricted class abstraction in a subclass relationship. (Contributed by Glauco Siliprandi, 26-Jun-2021)

Ref Expression
Hypothesis rabssf.1 ⊢ Ⅎ 𝑥 𝐵
Assertion rabssf ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ⊆ 𝐵 ↔ ∀ 𝑥 ∈ 𝐴 ( 𝜑 → 𝑥 ∈ 𝐵 ) )

Proof

Step Hyp Ref Expression
1 rabssf.1 ⊢ Ⅎ 𝑥 𝐵
2 df-rab ⊢ { 𝑥 ∈ 𝐴 ∣ 𝜑 } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) }
3 2 sseq1i ⊢ ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ⊆ 𝐵 ↔ { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ⊆ 𝐵 )
4 1 abssf ⊢ ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ⊆ 𝐵 ↔ ∀ 𝑥 ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) → 𝑥 ∈ 𝐵 ) )
5 impexp ⊢ ( ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) → 𝑥 ∈ 𝐵 ) ↔ ( 𝑥 ∈ 𝐴 → ( 𝜑 → 𝑥 ∈ 𝐵 ) ) )
6 5 albii ⊢ ( ∀ 𝑥 ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) → 𝑥 ∈ 𝐵 ) ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → ( 𝜑 → 𝑥 ∈ 𝐵 ) ) )
7 df-ral ⊢ ( ∀ 𝑥 ∈ 𝐴 ( 𝜑 → 𝑥 ∈ 𝐵 ) ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → ( 𝜑 → 𝑥 ∈ 𝐵 ) ) )
8 6 7 bitr4i ⊢ ( ∀ 𝑥 ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) → 𝑥 ∈ 𝐵 ) ↔ ∀ 𝑥 ∈ 𝐴 ( 𝜑 → 𝑥 ∈ 𝐵 ) )
9 3 4 8 3bitri ⊢ ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ⊆ 𝐵 ↔ ∀ 𝑥 ∈ 𝐴 ( 𝜑 → 𝑥 ∈ 𝐵 ) )