Theorem ssrabf 45119

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 3437	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
2	1	sseq2i 4013	. 2 ⊢ (𝐵 ⊆ {𝑥 ∈ 𝐴 ∣ 𝜑} ↔ 𝐵 ⊆ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)})
3		ssrabf.1	. . 3 ⊢ Ⅎ𝑥𝐵
4	3	ssabf 45105	. 2 ⊢ (𝐵 ⊆ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 ∧ 𝜑)))
5		ssrabf.2	. . . . 5 ⊢ Ⅎ𝑥𝐴
6	3, 5	dfss3f 3975	. . . 4 ⊢ (𝐵 ⊆ 𝐴 ↔ ∀𝑥 ∈ 𝐵 𝑥 ∈ 𝐴)
7	6	anbi1i 624	. . 3 ⊢ ((𝐵 ⊆ 𝐴 ∧ ∀𝑥 ∈ 𝐵 𝜑) ↔ (∀𝑥 ∈ 𝐵 𝑥 ∈ 𝐴 ∧ ∀𝑥 ∈ 𝐵 𝜑))
8		r19.26 3111	. . 3 ⊢ (∀𝑥 ∈ 𝐵 (𝑥 ∈ 𝐴 ∧ 𝜑) ↔ (∀𝑥 ∈ 𝐵 𝑥 ∈ 𝐴 ∧ ∀𝑥 ∈ 𝐵 𝜑))
9		df-ral 3062	. . 3 ⊢ (∀𝑥 ∈ 𝐵 (𝑥 ∈ 𝐴 ∧ 𝜑) ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 ∧ 𝜑)))
10	7, 8, 9	3bitr2ri 300	. 2 ⊢ (∀𝑥(𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 ∧ 𝜑)) ↔ (𝐵 ⊆ 𝐴 ∧ ∀𝑥 ∈ 𝐵 𝜑))
11	2, 4, 10	3bitri 297	1 ⊢ (𝐵 ⊆ {𝑥 ∈ 𝐴 ∣ 𝜑} ↔ (𝐵 ⊆ 𝐴 ∧ ∀𝑥 ∈ 𝐵 𝜑))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395  ∀wal 1538   ∈ wcel 2108  {cab 2714  Ⅎwnfc 2890  ∀wral 3061  {crab 3436   ⊆ wss 3951

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-10 2141  ax-11 2157  ax-12 2177  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1543  df-ex 1780  df-nf 1784  df-sb 2065  df-clab 2715  df-cleq 2729  df-clel 2816  df-nfc 2892  df-ral 3062  df-rab 3437  df-ss 3968

This theorem is referenced by:  ssrabdf  45120  supminfxr2  45480  pimgtmnf2  46729  smfmullem4  46809  smflimsuplem7  46841