Theorem abss 4038

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 2872	. . 3 ⊢ {𝑥 ∣ 𝑥 ∈ 𝐴} = 𝐴
2	1	sseq2i 3988	. 2 ⊢ ({𝑥 ∣ 𝜑} ⊆ {𝑥 ∣ 𝑥 ∈ 𝐴} ↔ {𝑥 ∣ 𝜑} ⊆ 𝐴)
3		ss2ab 4037	. 2 ⊢ ({𝑥 ∣ 𝜑} ⊆ {𝑥 ∣ 𝑥 ∈ 𝐴} ↔ ∀𝑥(𝜑 → 𝑥 ∈ 𝐴))
4	2, 3	bitr3i 277	1 ⊢ ({𝑥 ∣ 𝜑} ⊆ 𝐴 ↔ ∀𝑥(𝜑 → 𝑥 ∈ 𝐴))

Syntax hints:   → wi 4   ↔ wb 206  ∀wal 1538   ∈ wcel 2108  {cab 2713   ⊆ wss 3926

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 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-10 2141  ax-11 2157  ax-12 2177  ax-ext 2707

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1543  df-ex 1780  df-nf 1784  df-sb 2065  df-clab 2714  df-cleq 2727  df-clel 2809  df-nfc 2885  df-ss 3943

This theorem is referenced by:  abssdvOLD  4044  rabss  4047  uniiunlem  4062  iunssf  5020  iunss  5021  abexd  5295  abex  5296  moabex  5434  reliun  5795  axdc2lem  10462  mptelee  28874  fpwrelmap  32710  hoidmvlelem1  46624