Theorem abss 4058

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 2869	. . 3 ⊢ {𝑥 ∣ 𝑥 ∈ 𝐴} = 𝐴
2	1	sseq2i 4012	. 2 ⊢ ({𝑥 ∣ 𝜑} ⊆ {𝑥 ∣ 𝑥 ∈ 𝐴} ↔ {𝑥 ∣ 𝜑} ⊆ 𝐴)
3		ss2ab 4057	. 2 ⊢ ({𝑥 ∣ 𝜑} ⊆ {𝑥 ∣ 𝑥 ∈ 𝐴} ↔ ∀𝑥(𝜑 → 𝑥 ∈ 𝐴))
4	2, 3	bitr3i 276	1 ⊢ ({𝑥 ∣ 𝜑} ⊆ 𝐴 ↔ ∀𝑥(𝜑 → 𝑥 ∈ 𝐴))

Syntax hints:   → wi 4   ↔ wb 205  ∀wal 1537   ∈ wcel 2104  {cab 2707   ⊆ wss 3949

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1911  ax-6 1969  ax-7 2009  ax-8 2106  ax-9 2114  ax-10 2135  ax-11 2152  ax-12 2169  ax-ext 2701

This theorem depends on definitions:  df-bi 206  df-an 395  df-or 844  df-tru 1542  df-ex 1780  df-nf 1784  df-sb 2066  df-clab 2708  df-cleq 2722  df-clel 2808  df-nfc 2883  df-v 3474  df-in 3956  df-ss 3966

This theorem is referenced by:  abssdvOLD  4067  rabss  4070  uniiunlem  4085  iunssf  5048  iunss  5049  moabex  5460  reliun  5817  axdc2lem  10447  mptelee  28418  fpwrelmap  32223  hoidmvlelem1  45611