Theorem abss 3951

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 2875	. . 3 ⊢ {𝑥 ∣ 𝑥 ∈ 𝐴} = 𝐴
2	1	sseq2i 3907	. 2 ⊢ ({𝑥 ∣ 𝜑} ⊆ {𝑥 ∣ 𝑥 ∈ 𝐴} ↔ {𝑥 ∣ 𝜑} ⊆ 𝐴)
3		ss2ab 3950	. 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 1540   ∈ wcel 2114  {cab 2717   ⊆ wss 3844

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1975  ax-7 2020  ax-8 2116  ax-9 2124  ax-10 2145  ax-11 2162  ax-12 2179  ax-ext 2711

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 847  df-tru 1545  df-ex 1787  df-nf 1791  df-sb 2075  df-clab 2718  df-cleq 2731  df-clel 2812  df-nfc 2882  df-v 3401  df-in 3851  df-ss 3861

This theorem is referenced by:  abssdv  3959  rabss  3962  uniiunlem  3976  iunssf  4931  iunss  4932  moabex  5318  reliun  5661  axdc2lem  9951  mptelee  26844  fpwrelmap  30646  ss2iundf  40836  hoidmvlelem1  43698