MPE Home Metamath Proof Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >  abss Structured version   Unicode version

Theorem abss 3555
Description: Class abstraction in a subclass relationship. (Contributed by NM, 16-Aug-2006.)
Assertion
Ref Expression
abss  |-  ( { x  |  ph }  C_  A  <->  A. x ( ph  ->  x  e.  A ) )
Distinct variable group:    x, A
Allowed substitution hint:    ph( x)

Proof of Theorem abss
StepHypRef Expression
1 abid2 2594 . . 3  |-  { x  |  x  e.  A }  =  A
21sseq2i 3514 . 2  |-  ( { x  |  ph }  C_ 
{ x  |  x  e.  A }  <->  { x  |  ph }  C_  A
)
3 ss2ab 3554 . 2  |-  ( { x  |  ph }  C_ 
{ x  |  x  e.  A }  <->  A. x
( ph  ->  x  e.  A ) )
42, 3bitr3i 251 1  |-  ( { x  |  ph }  C_  A  <->  A. x ( ph  ->  x  e.  A ) )
Colors of variables: wff setvar class
Syntax hints:    -> wi 4    <-> wb 184   A.wal 1396    e. wcel 1823   {cab 2439    C_ wss 3461
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1623  ax-4 1636  ax-5 1709  ax-6 1752  ax-7 1795  ax-10 1842  ax-11 1847  ax-12 1859  ax-13 2004  ax-ext 2432
This theorem depends on definitions:  df-bi 185  df-or 368  df-an 369  df-tru 1401  df-ex 1618  df-nf 1622  df-sb 1745  df-clab 2440  df-cleq 2446  df-clel 2449  df-nfc 2604  df-in 3468  df-ss 3475
This theorem is referenced by:  abssdv  3560  rabss  3563  uniiunlem  3574  iunss  4356  moabex  4696  reliun  5111  axdc2lem  8819  mptelee  24403  fpwrelmap  27790
  Copyright terms: Public domain W3C validator