Theorem ssrab 3012

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 2309	. . 3 ⊢ {x ∈ A ∣ φ} = {x ∣ (x ∈ A ∧ φ)}
2	1	sseq2i 2964	. 2 ⊢ (B ⊆ {x ∈ A ∣ φ} ↔ B ⊆ {x ∣ (x ∈ A ∧ φ)})
3		ssab 3004	. 2 ⊢ (B ⊆ {x ∣ (x ∈ A ∧ φ)} ↔ ∀x(x ∈ B → (x ∈ A ∧ φ)))
4		dfss3 2929	. . . 4 ⊢ (B ⊆ A ↔ ∀x ∈ B x ∈ A)
5	4	anbi1i 431	. . 3 ⊢ ((B ⊆ A ∧ ∀x ∈ B φ) ↔ (∀x ∈ B x ∈ A ∧ ∀x ∈ B φ))
6		r19.26 2435	. . 3 ⊢ (∀x ∈ B (x ∈ A ∧ φ) ↔ (∀x ∈ B x ∈ A ∧ ∀x ∈ B φ))
7		df-ral 2305	. . 3 ⊢ (∀x ∈ B (x ∈ A ∧ φ) ↔ ∀x(x ∈ B → (x ∈ A ∧ φ)))
8	5, 6, 7	3bitr2ri 198	. 2 ⊢ (∀x(x ∈ B → (x ∈ A ∧ φ)) ↔ (B ⊆ A ∧ ∀x ∈ B φ))
9	2, 3, 8	3bitri 195	1 ⊢ (B ⊆ {x ∈ A ∣ φ} ↔ (B ⊆ A ∧ ∀x ∈ B φ))

Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98  ∀wal 1240   ∈ wcel 1390  {cab 2023  ∀wral 2300  {crab 2304   ⊆ wss 2911

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bnd 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019

This theorem depends on definitions:  df-bi 110  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-ral 2305  df-rab 2309  df-in 2918  df-ss 2925

This theorem is referenced by:  ssrabdv  3013