Theorem abss 4018

Description: Class abstraction in a subclass relationship. (Contributed by NM, 16-Aug-2006.)

Assertion

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

Distinct variable group:   𝑥,𝐴

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem abss

Step	Hyp	Ref	Expression
1		abid2 2902	. . 3 ⊢ {𝑥 ∣ 𝑥 ∈ 𝐴} = 𝐴
2	1	sseq2i 3968	. 2 ⊢ ({𝑥 ∣ 𝜑} ⊆ {𝑥 ∣ 𝑥 ∈ 𝐴} ↔ {𝑥 ∣ 𝜑} ⊆ 𝐴)
3		ss2ab 4017	. 2 ⊢ ({𝑥 ∣ 𝜑} ⊆ {𝑥 ∣ 𝑥 ∈ 𝐴} ↔ ∀𝑥(𝜑 → 𝑥 ∈ 𝐴))
4	2, 3	bitr3i 280	1 ⊢ ({𝑥 ∣ 𝜑} ⊆ 𝐴 ↔ ∀𝑥(𝜑 → 𝑥 ∈ 𝐴))

Syntax hints:   → wi 4   ↔ wb 209  ∀wal 1561   ∈ wcel 2145  {cab 2743   ⊆ wss 3907

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1818  ax-4 1832  ax-5 1933  ax-6 1990  ax-7 2031  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2215  ax-ext 2737

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1566  df-ex 1803  df-nf 1807  df-sb 2094  df-clab 2744  df-cleq 2757  df-clel 2840  df-nfc 2914  df-ss 3924

This theorem is referenced by:  rabss  4026  uniiunlem  4043  iunssfOLD  5004  iunssOLD  5006  moabexOLD  5431  mpteleeOLD  29154