Theorem ssintrab 4930

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 3413	. . . 4 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
2	1	inteqi 4910	. . 3 ⊢ ∩ {𝑥 ∈ 𝐵 ∣ 𝜑} = ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
3	2	sseq2i 3938	. 2 ⊢ (𝐴 ⊆ ∩ {𝑥 ∈ 𝐵 ∣ 𝜑} ↔ 𝐴 ⊆ ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)})
4		impexp 456	. . . 4 ⊢ (((𝑥 ∈ 𝐵 ∧ 𝜑) → 𝐴 ⊆ 𝑥) ↔ (𝑥 ∈ 𝐵 → (𝜑 → 𝐴 ⊆ 𝑥)))
5	4	albii 1851	. . 3 ⊢ (∀𝑥((𝑥 ∈ 𝐵 ∧ 𝜑) → 𝐴 ⊆ 𝑥) ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝜑 → 𝐴 ⊆ 𝑥)))
6		ssintab 4924	. . 3 ⊢ (𝐴 ⊆ ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)} ↔ ∀𝑥((𝑥 ∈ 𝐵 ∧ 𝜑) → 𝐴 ⊆ 𝑥))
7		df-ral 3077	. . 3 ⊢ (∀𝑥 ∈ 𝐵 (𝜑 → 𝐴 ⊆ 𝑥) ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝜑 → 𝐴 ⊆ 𝑥)))
8	5, 6, 7	3bitr4i 306	. 2 ⊢ (𝐴 ⊆ ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)} ↔ ∀𝑥 ∈ 𝐵 (𝜑 → 𝐴 ⊆ 𝑥))
9	3, 8	bitri 278	1 ⊢ (𝐴 ⊆ ∩ {𝑥 ∈ 𝐵 ∣ 𝜑} ↔ ∀𝑥 ∈ 𝐵 (𝜑 → 𝐴 ⊆ 𝑥))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1567   ∈ wcel 2144  {cab 2738  ∀wral 3076  {crab 3412   ⊆ wss 3896  ∩ cint 4906

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-10 2177  ax-11 2193  ax-12 2212  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-ex 1812  df-nf 1816  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-ss 3897  df-int 4907

This theorem is used by:  knatar  7371  harval2  10085  pwfseqlem3  10752  elrgspnlem4  33856  ldgenpisyslem3  34848  topjoin  37249