Theorem ssintrab 4995

Description: Subclass of the intersection of a restricted class abstraction. (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 3444	. . . 4 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
2	1	inteqi 4974	. . 3 ⊢ ∩ {𝑥 ∈ 𝐵 ∣ 𝜑} = ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
3	2	sseq2i 4038	. 2 ⊢ (𝐴 ⊆ ∩ {𝑥 ∈ 𝐵 ∣ 𝜑} ↔ 𝐴 ⊆ ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)})
4		impexp 450	. . . 4 ⊢ (((𝑥 ∈ 𝐵 ∧ 𝜑) → 𝐴 ⊆ 𝑥) ↔ (𝑥 ∈ 𝐵 → (𝜑 → 𝐴 ⊆ 𝑥)))
5	4	albii 1817	. . 3 ⊢ (∀𝑥((𝑥 ∈ 𝐵 ∧ 𝜑) → 𝐴 ⊆ 𝑥) ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝜑 → 𝐴 ⊆ 𝑥)))
6		ssintab 4989	. . 3 ⊢ (𝐴 ⊆ ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)} ↔ ∀𝑥((𝑥 ∈ 𝐵 ∧ 𝜑) → 𝐴 ⊆ 𝑥))
7		df-ral 3068	. . 3 ⊢ (∀𝑥 ∈ 𝐵 (𝜑 → 𝐴 ⊆ 𝑥) ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝜑 → 𝐴 ⊆ 𝑥)))
8	5, 6, 7	3bitr4i 303	. 2 ⊢ (𝐴 ⊆ ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)} ↔ ∀𝑥 ∈ 𝐵 (𝜑 → 𝐴 ⊆ 𝑥))
9	3, 8	bitri 275	1 ⊢ (𝐴 ⊆ ∩ {𝑥 ∈ 𝐵 ∣ 𝜑} ↔ ∀𝑥 ∈ 𝐵 (𝜑 → 𝐴 ⊆ 𝑥))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395  ∀wal 1535   ∈ wcel 2108  {cab 2717  ∀wral 3067  {crab 3443   ⊆ wss 3976  ∩ cint 4970

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-10 2141  ax-11 2158  ax-12 2178  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-tru 1540  df-ex 1778  df-nf 1782  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-ral 3068  df-rex 3077  df-rab 3444  df-v 3490  df-ss 3993  df-int 4971

This theorem is referenced by:  knatar  7393  harval2  10066  pwfseqlem3  10729  ldgenpisyslem3  34129  topjoin  36331