Theorem ssab 3863

Description: Subclass of a class abstraction. (Contributed by NM, 16-Aug-2006.)

Assertion

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

Distinct variable group:   𝑥,𝐴

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem ssab

Step	Hyp	Ref	Expression
1		abid2 2925	. . 3 ⊢ {𝑥 ∣ 𝑥 ∈ 𝐴} = 𝐴
2	1	sseq1i 3820	. 2 ⊢ ({𝑥 ∣ 𝑥 ∈ 𝐴} ⊆ {𝑥 ∣ 𝜑} ↔ 𝐴 ⊆ {𝑥 ∣ 𝜑})
3		ss2ab 3861	. 2 ⊢ ({𝑥 ∣ 𝑥 ∈ 𝐴} ⊆ {𝑥 ∣ 𝜑} ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝜑))
4	2, 3	bitr3i 268	1 ⊢ (𝐴 ⊆ {𝑥 ∣ 𝜑} ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝜑))

Syntax hints:   → wi 4   ↔ wb 197  ∀wal 1635   ∈ wcel 2155  {cab 2788   ⊆ wss 3763

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1877  ax-4 1894  ax-5 2001  ax-6 2067  ax-7 2103  ax-9 2164  ax-10 2184  ax-11 2200  ax-12 2213  ax-13 2419  ax-ext 2781

This theorem depends on definitions:  df-bi 198  df-an 385  df-or 866  df-tru 1641  df-ex 1860  df-nf 1864  df-sb 2060  df-clab 2789  df-cleq 2795  df-clel 2798  df-nfc 2933  df-in 3770  df-ss 3777

This theorem is referenced by:  ssabral  3864  ssrab  3871  wdomd  8719  ixpiunwdom  8729  lidldvgen  19458  prdsxmslem2  22541  ballotlem2  30869