Theorem abss 3633

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 2731	. . 3 ⊢ {𝑥 ∣ 𝑥 ∈ 𝐴} = 𝐴
2	1	sseq2i 3592	. 2 ⊢ ({𝑥 ∣ 𝜑} ⊆ {𝑥 ∣ 𝑥 ∈ 𝐴} ↔ {𝑥 ∣ 𝜑} ⊆ 𝐴)
3		ss2ab 3632	. 2 ⊢ ({𝑥 ∣ 𝜑} ⊆ {𝑥 ∣ 𝑥 ∈ 𝐴} ↔ ∀𝑥(𝜑 → 𝑥 ∈ 𝐴))
4	2, 3	bitr3i 264	1 ⊢ ({𝑥 ∣ 𝜑} ⊆ 𝐴 ↔ ∀𝑥(𝜑 → 𝑥 ∈ 𝐴))

Syntax hints:   → wi 4   ↔ wb 194  ∀wal 1472   ∈ wcel 1976  {cab 2595   ⊆ wss 3539

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1712  ax-4 1727  ax-5 1826  ax-6 1874  ax-7 1921  ax-10 2005  ax-11 2020  ax-12 2033  ax-13 2233  ax-ext 2589

This theorem depends on definitions:  df-bi 195  df-or 383  df-an 384  df-tru 1477  df-ex 1695  df-nf 1700  df-sb 1867  df-clab 2596  df-cleq 2602  df-clel 2605  df-nfc 2739  df-in 3546  df-ss 3553

This theorem is referenced by:  abssdv  3638  rabss  3641  uniiunlem  3652  iunss  4491  moabex  4847  reliun  5150  axdc2lem  9130  mptelee  25520  fpwrelmap  28689  ss2iundf  36753  iunssf  38073  hoidmvlelem1  39268