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

Theorem abss 3255
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 2413 . . 3  |-  { x  |  x  e.  A }  =  A
21sseq2i 3216 . 2  |-  ( { x  |  ph }  C_ 
{ x  |  x  e.  A }  <->  { x  |  ph }  C_  A
)
3 ss2ab 3254 . 2  |-  ( { x  |  ph }  C_ 
{ x  |  x  e.  A }  <->  A. x
( ph  ->  x  e.  A ) )
42, 3bitr3i 242 1  |-  ( { x  |  ph }  C_  A  <->  A. x ( ph  ->  x  e.  A ) )
Colors of variables: wff set class
Syntax hints:    -> wi 4    <-> wb 176   A.wal 1530    e. wcel 1696   {cab 2282    C_ wss 3165
This theorem is referenced by:  abssdv  3260  rabss  3263  uniiunlem  3273  iunss  3959  moabex  4248  reliun  4822  axdc2lem  8090  mptelee  24595  qusp  25645
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1536  ax-5 1547  ax-17 1606  ax-9 1644  ax-8 1661  ax-6 1715  ax-7 1720  ax-11 1727  ax-12 1878  ax-ext 2277
This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1310  df-ex 1532  df-nf 1535  df-sb 1639  df-clab 2283  df-cleq 2289  df-clel 2292  df-nfc 2421  df-in 3172  df-ss 3179
  Copyright terms: Public domain W3C validator