Theorem ssintrab 3950

Description: Subclass of the intersection of a restricted class builder. (Contributed by NM, 30-Jan-2015.)

Assertion

Ref	Expression
ssintrab	⊢ (A ⊆ ∩{x ∈ B ∣ φ} ↔ ∀x ∈ B (φ → A ⊆ x))

Distinct variable group:   x,A

Allowed substitution hints:   φ(x)   B(x)

Proof of Theorem ssintrab

Step	Hyp	Ref	Expression
1		df-rab 2624	. . . 4 ⊢ {x ∈ B ∣ φ} = {x ∣ (x ∈ B ∧ φ)}
2	1	inteqi 3931	. . 3 ⊢ ∩{x ∈ B ∣ φ} = ∩{x ∣ (x ∈ B ∧ φ)}
3	2	sseq2i 3297	. 2 ⊢ (A ⊆ ∩{x ∈ B ∣ φ} ↔ A ⊆ ∩{x ∣ (x ∈ B ∧ φ)})
4		impexp 433	. . . 4 ⊢ (((x ∈ B ∧ φ) → A ⊆ x) ↔ (x ∈ B → (φ → A ⊆ x)))
5	4	albii 1566	. . 3 ⊢ (∀x((x ∈ B ∧ φ) → A ⊆ x) ↔ ∀x(x ∈ B → (φ → A ⊆ x)))
6		ssintab 3944	. . 3 ⊢ (A ⊆ ∩{x ∣ (x ∈ B ∧ φ)} ↔ ∀x((x ∈ B ∧ φ) → A ⊆ x))
7		df-ral 2620	. . 3 ⊢ (∀x ∈ B (φ → A ⊆ x) ↔ ∀x(x ∈ B → (φ → A ⊆ x)))
8	5, 6, 7	3bitr4i 268	. 2 ⊢ (A ⊆ ∩{x ∣ (x ∈ B ∧ φ)} ↔ ∀x ∈ B (φ → A ⊆ x))
9	3, 8	bitri 240	1 ⊢ (A ⊆ ∩{x ∈ B ∣ φ} ↔ ∀x ∈ B (φ → A ⊆ x))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∧ wa 358  ∀wal 1540   ∈ wcel 1710  {cab 2339  ∀wral 2615  {crab 2619   ⊆ wss 3258  ∩cint 3927

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  df-int 3928

This theorem is used by: (None)