Theorem ssrabf 45544

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 3390	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
2	1	sseq2i 3951	. 2 ⊢ (𝐵 ⊆ {𝑥 ∈ 𝐴 ∣ 𝜑} ↔ 𝐵 ⊆ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)})
3		ssrabf.1	. . 3 ⊢ Ⅎ𝑥𝐵
4	3	ssabf 45530	. 2 ⊢ (𝐵 ⊆ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 ∧ 𝜑)))
5		ssrabf.2	. . . . 5 ⊢ Ⅎ𝑥𝐴
6	3, 5	dfss3f 3913	. . . 4 ⊢ (𝐵 ⊆ 𝐴 ↔ ∀𝑥 ∈ 𝐵 𝑥 ∈ 𝐴)
7	6	anbi1i 625	. . 3 ⊢ ((𝐵 ⊆ 𝐴 ∧ ∀𝑥 ∈ 𝐵 𝜑) ↔ (∀𝑥 ∈ 𝐵 𝑥 ∈ 𝐴 ∧ ∀𝑥 ∈ 𝐵 𝜑))
8		r19.26 3097	. . 3 ⊢ (∀𝑥 ∈ 𝐵 (𝑥 ∈ 𝐴 ∧ 𝜑) ↔ (∀𝑥 ∈ 𝐵 𝑥 ∈ 𝐴 ∧ ∀𝑥 ∈ 𝐵 𝜑))
9		df-ral 3052	. . 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 1540   ∈ wcel 2114  {cab 2714  Ⅎwnfc 2883  ∀wral 3051  {crab 3389   ⊆ wss 3889

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-10 2147  ax-11 2163  ax-12 2185  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1545  df-ex 1782  df-nf 1786  df-sb 2069  df-clab 2715  df-cleq 2728  df-clel 2811  df-nfc 2885  df-ral 3052  df-rab 3390  df-ss 3906

This theorem is referenced by:  ssrabdf  45545  supminfxr2  45897  pimgtmnf2  47142  smfmullem4  47222  smflimsuplem7  47254