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Metamath Proof Explorer

Theorem ssintrab

Description: Subclass of the intersection of a restricted class abstraction. (Contributed by NM, 30-Jan-2015)

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

Proof

Step Hyp Ref Expression
1 df-rab ⊢ { 𝑥 ∈ 𝐵 ∣ 𝜑 } = { 𝑥 ∣ ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) }
2 1 inteqi ⊢ ∩ { 𝑥 ∈ 𝐵 ∣ 𝜑 } = ∩ { 𝑥 ∣ ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) }
3 2 sseq2i ⊢ ( 𝐴 ⊆ ∩ { 𝑥 ∈ 𝐵 ∣ 𝜑 } ↔ 𝐴 ⊆ ∩ { 𝑥 ∣ ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) } )
4 impexp ⊢ ( ( ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) → 𝐴 ⊆ 𝑥 ) ↔ ( 𝑥 ∈ 𝐵 → ( 𝜑 → 𝐴 ⊆ 𝑥 ) ) )
5 4 albii ⊢ ( ∀ 𝑥 ( ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) → 𝐴 ⊆ 𝑥 ) ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐵 → ( 𝜑 → 𝐴 ⊆ 𝑥 ) ) )
6 ssintab ⊢ ( 𝐴 ⊆ ∩ { 𝑥 ∣ ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) } ↔ ∀ 𝑥 ( ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) → 𝐴 ⊆ 𝑥 ) )
7 df-ral ⊢ ( ∀ 𝑥 ∈ 𝐵 ( 𝜑 → 𝐴 ⊆ 𝑥 ) ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐵 → ( 𝜑 → 𝐴 ⊆ 𝑥 ) ) )
8 5 6 7 3bitr4i ⊢ ( 𝐴 ⊆ ∩ { 𝑥 ∣ ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) } ↔ ∀ 𝑥 ∈ 𝐵 ( 𝜑 → 𝐴 ⊆ 𝑥 ) )
9 3 8 bitri ⊢ ( 𝐴 ⊆ ∩ { 𝑥 ∈ 𝐵 ∣ 𝜑 } ↔ ∀ 𝑥 ∈ 𝐵 ( 𝜑 → 𝐴 ⊆ 𝑥 ) )