Theorem ssrab 3345

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 2624	. . 3 ⊢ {x ∈ A ∣ φ} = {x ∣ (x ∈ A ∧ φ)}
2	1	sseq2i 3297	. 2 ⊢ (B ⊆ {x ∈ A ∣ φ} ↔ B ⊆ {x ∣ (x ∈ A ∧ φ)})
3		ssab 3337	. 2 ⊢ (B ⊆ {x ∣ (x ∈ A ∧ φ)} ↔ ∀x(x ∈ B → (x ∈ A ∧ φ)))
4		dfss3 3264	. . . 4 ⊢ (B ⊆ A ↔ ∀x ∈ B x ∈ A)
5	4	anbi1i 676	. . 3 ⊢ ((B ⊆ A ∧ ∀x ∈ B φ) ↔ (∀x ∈ B x ∈ A ∧ ∀x ∈ B φ))
6		r19.26 2747	. . 3 ⊢ (∀x ∈ B (x ∈ A ∧ φ) ↔ (∀x ∈ B x ∈ A ∧ ∀x ∈ B φ))
7		df-ral 2620	. . 3 ⊢ (∀x ∈ B (x ∈ A ∧ φ) ↔ ∀x(x ∈ B → (x ∈ A ∧ φ)))
8	5, 6, 7	3bitr2ri 265	. 2 ⊢ (∀x(x ∈ B → (x ∈ A ∧ φ)) ↔ (B ⊆ A ∧ ∀x ∈ B φ))
9	2, 3, 8	3bitri 262	1 ⊢ (B ⊆ {x ∈ A ∣ φ} ↔ (B ⊆ A ∧ ∀x ∈ B φ))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∧ wa 358  ∀wal 1540   ∈ wcel 1710  {cab 2339  ∀wral 2615  {crab 2619   ⊆ wss 3258

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-ral 2620  df-rab 2624  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-ss 3260

This theorem is used by:  ssrabdv  3346