Theorem abss 3336

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 2471	. . 3 ⊢ {x ∣ x ∈ A} = A
2	1	sseq2i 3297	. 2 ⊢ ({x ∣ φ} ⊆ {x ∣ x ∈ A} ↔ {x ∣ φ} ⊆ A)
3		ss2ab 3335	. 2 ⊢ ({x ∣ φ} ⊆ {x ∣ x ∈ A} ↔ ∀x(φ → x ∈ A))
4	2, 3	bitr3i 242	1 ⊢ ({x ∣ φ} ⊆ A ↔ ∀x(φ → x ∈ A))

Syntax hints:   → wi 4   ↔ wb 176  ∀wal 1540   ∈ wcel 1710  {cab 2339   ⊆ wss 3258

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-ss 3260

This theorem is referenced by:  abssdv  3341  rabss  3344  uniiunlem  3354  iunss  4008