Theorem ssrabf 40056

Description: Subclass of a restricted class abstraction. (Contributed by Glauco Siliprandi, 26-Jun-2021.)

Hypotheses

Ref	Expression
ssrabf.1	⊢ Ⅎ𝑥𝐵
ssrabf.2	⊢ Ⅎ𝑥𝐴

Assertion

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

Proof of Theorem ssrabf

Step	Hyp	Ref	Expression
1		df-rab 3098	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
2	1	sseq2i 3826	. 2 ⊢ (𝐵 ⊆ {𝑥 ∈ 𝐴 ∣ 𝜑} ↔ 𝐵 ⊆ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)})
3		ssrabf.1	. . 3 ⊢ Ⅎ𝑥𝐵
4	3	ssabf 40039	. 2 ⊢ (𝐵 ⊆ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 ∧ 𝜑)))
5		ssrabf.2	. . . . 5 ⊢ Ⅎ𝑥𝐴
6	3, 5	dfss3f 3790	. . . 4 ⊢ (𝐵 ⊆ 𝐴 ↔ ∀𝑥 ∈ 𝐵 𝑥 ∈ 𝐴)
7	6	anbi1i 618	. . 3 ⊢ ((𝐵 ⊆ 𝐴 ∧ ∀𝑥 ∈ 𝐵 𝜑) ↔ (∀𝑥 ∈ 𝐵 𝑥 ∈ 𝐴 ∧ ∀𝑥 ∈ 𝐵 𝜑))
8		r19.26 3245	. . 3 ⊢ (∀𝑥 ∈ 𝐵 (𝑥 ∈ 𝐴 ∧ 𝜑) ↔ (∀𝑥 ∈ 𝐵 𝑥 ∈ 𝐴 ∧ ∀𝑥 ∈ 𝐵 𝜑))
9		df-ral 3094	. . 3 ⊢ (∀𝑥 ∈ 𝐵 (𝑥 ∈ 𝐴 ∧ 𝜑) ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 ∧ 𝜑)))
10	7, 8, 9	3bitr2ri 292	. 2 ⊢ (∀𝑥(𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 ∧ 𝜑)) ↔ (𝐵 ⊆ 𝐴 ∧ ∀𝑥 ∈ 𝐵 𝜑))
11	2, 4, 10	3bitri 289	1 ⊢ (𝐵 ⊆ {𝑥 ∈ 𝐴 ∣ 𝜑} ↔ (𝐵 ⊆ 𝐴 ∧ ∀𝑥 ∈ 𝐵 𝜑))

Syntax hints:   → wi 4   ↔ wb 198   ∧ wa 385  ∀wal 1651   ∈ wcel 2157  {cab 2785  Ⅎwnfc 2928  ∀wral 3089  {crab 3093   ⊆ wss 3769

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1891  ax-4 1905  ax-5 2006  ax-6 2072  ax-7 2107  ax-9 2166  ax-10 2185  ax-11 2200  ax-12 2213  ax-13 2377  ax-ext 2777

This theorem depends on definitions:  df-bi 199  df-an 386  df-or 875  df-tru 1657  df-ex 1876  df-nf 1880  df-sb 2065  df-clab 2786  df-cleq 2792  df-clel 2795  df-nfc 2930  df-ral 3094  df-rab 3098  df-in 3776  df-ss 3783

This theorem is referenced by:  supminfxr2  40442  pimgtmnf2  41670  smfmullem4  41747  smflimsuplem7  41778