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Metamath Proof Explorer

Theorem ssrab

Description: Subclass of a restricted class abstraction. (Contributed by NM, 16-Aug-2006)

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

Proof

Step Hyp Ref Expression
1 df-rab ⊢ { 𝑥 ∈ 𝐴 ∣ 𝜑 } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) }
2 1 sseq2i ⊢ ( 𝐵 ⊆ { 𝑥 ∈ 𝐴 ∣ 𝜑 } ↔ 𝐵 ⊆ { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } )
3 ssab ⊢ ( 𝐵 ⊆ { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐵 → ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) )
4 dfss3 ⊢ ( 𝐵 ⊆ 𝐴 ↔ ∀ 𝑥 ∈ 𝐵 𝑥 ∈ 𝐴 )
5 4 anbi1i ⊢ ( ( 𝐵 ⊆ 𝐴 ∧ ∀ 𝑥 ∈ 𝐵 𝜑 ) ↔ ( ∀ 𝑥 ∈ 𝐵 𝑥 ∈ 𝐴 ∧ ∀ 𝑥 ∈ 𝐵 𝜑 ) )
6 r19.26 ⊢ ( ∀ 𝑥 ∈ 𝐵 ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ↔ ( ∀ 𝑥 ∈ 𝐵 𝑥 ∈ 𝐴 ∧ ∀ 𝑥 ∈ 𝐵 𝜑 ) )
7 df-ral ⊢ ( ∀ 𝑥 ∈ 𝐵 ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐵 → ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) )
8 5 6 7 3bitr2ri ⊢ ( ∀ 𝑥 ( 𝑥 ∈ 𝐵 → ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) ↔ ( 𝐵 ⊆ 𝐴 ∧ ∀ 𝑥 ∈ 𝐵 𝜑 ) )
9 2 3 8 3bitri ⊢ ( 𝐵 ⊆ { 𝑥 ∈ 𝐴 ∣ 𝜑 } ↔ ( 𝐵 ⊆ 𝐴 ∧ ∀ 𝑥 ∈ 𝐵 𝜑 ) )