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Metamath Proof Explorer

Theorem rabss

Description: Restricted class abstraction in a subclass relationship. (Contributed by NM, 16-Aug-2006)

Ref Expression
Assertion rabss ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ⊆ 𝐵 ↔ ∀ 𝑥 ∈ 𝐴 ( 𝜑 → 𝑥 ∈ 𝐵 ) )

Proof

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