Theorem ssintrab 4938

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 3419	. . . 4 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
2	1	inteqi 4918	. . 3 ⊢ ∩ {𝑥 ∈ 𝐵 ∣ 𝜑} = ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
3	2	sseq2i 3967	. 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 1852	. . 3 ⊢ (∀𝑥((𝑥 ∈ 𝐵 ∧ 𝜑) → 𝐴 ⊆ 𝑥) ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝜑 → 𝐴 ⊆ 𝑥)))
6		ssintab 4932	. . 3 ⊢ (𝐴 ⊆ ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)} ↔ ∀𝑥((𝑥 ∈ 𝐵 ∧ 𝜑) → 𝐴 ⊆ 𝑥))
7		df-ral 3082	. . 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 1568   ∈ wcel 2146  {cab 2743  ∀wral 3081  {crab 3418   ⊆ wss 3906  ∩ cint 4914

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-10 2179  ax-11 2195  ax-12 2216  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-ral 3082  df-rex 3092  df-rab 3419  df-v 3459  df-ss 3923  df-int 4915

This theorem is used by:  knatar  7363  harval2  9995  pwfseqlem3  10656  elrgspnlem4  33605  ldgenpisyslem3  34596  topjoin  36909