Theorem ssrab 3315

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

Assertion

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

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem ssrab

Step	Hyp	Ref	Expression
1		df-rab 2529	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
2	1	sseq2i 3264	. 2 ⊢ (𝐵 ⊆ {𝑥 ∈ 𝐴 ∣ 𝜑} ↔ 𝐵 ⊆ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)})
3		ssab 3307	. 2 ⊢ (𝐵 ⊆ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 ∧ 𝜑)))
4		dfss3 3226	. . . 4 ⊢ (𝐵 ⊆ 𝐴 ↔ ∀𝑥 ∈ 𝐵 𝑥 ∈ 𝐴)
5	4	anbi1i 458	. . 3 ⊢ ((𝐵 ⊆ 𝐴 ∧ ∀𝑥 ∈ 𝐵 𝜑) ↔ (∀𝑥 ∈ 𝐵 𝑥 ∈ 𝐴 ∧ ∀𝑥 ∈ 𝐵 𝜑))
6		r19.26 2669	. . 3 ⊢ (∀𝑥 ∈ 𝐵 (𝑥 ∈ 𝐴 ∧ 𝜑) ↔ (∀𝑥 ∈ 𝐵 𝑥 ∈ 𝐴 ∧ ∀𝑥 ∈ 𝐵 𝜑))
7		df-ral 2525	. . 3 ⊢ (∀𝑥 ∈ 𝐵 (𝑥 ∈ 𝐴 ∧ 𝜑) ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 ∧ 𝜑)))
8	5, 6, 7	3bitr2ri 209	. 2 ⊢ (∀𝑥(𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 ∧ 𝜑)) ↔ (𝐵 ⊆ 𝐴 ∧ ∀𝑥 ∈ 𝐵 𝜑))
9	2, 3, 8	3bitri 206	1 ⊢ (𝐵 ⊆ {𝑥 ∈ 𝐴 ∣ 𝜑} ↔ (𝐵 ⊆ 𝐴 ∧ ∀𝑥 ∈ 𝐵 𝜑))

Syntax hints:   → wi 4   ∧ wa 104   ↔ wb 105  ∀wal 1396   ∈ wcel 2203  {cab 2218  ∀wral 2520  {crab 2524   ⊆ wss 3210

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2214

This theorem depends on definitions:  df-bi 117  df-nf 1510  df-sb 1812  df-clab 2219  df-cleq 2225  df-clel 2228  df-nfc 2373  df-ral 2525  df-rab 2529  df-in 3216  df-ss 3223

This theorem is referenced by:  ssrabdv  3316  frind  4472  epttop  14947