Theorem ab0 3570

Description: Empty class abstraction. (Contributed by SF, 5-Jan-2018.)

Assertion

Ref	Expression
ab0	|- ({x | ph} = (/) <-> A.x -. ph)

Proof of Theorem ab0

Step	Hyp	Ref	Expression
1		abn0 3569	. . 3 |- ({x | ph} =/= (/) <-> E.xph)
2		df-ne 2519	. . 3 |- ({x | ph} =/= (/) <-> -. {x | ph} = (/))
3		df-ex 1542	. . 3 |- (E.xph <-> -. A.x -. ph)
4	1, 2, 3	3bitr3i 266	. 2 |- (-. {x | ph} = (/) <-> -. A.x -. ph)
5	4	con4bii 288	1 |- ({x | ph} = (/) <-> A.x -. ph)

Syntax hints:  -. wn 3   <-> wb 176  A.wal 1540  E.wex 1541   = wceq 1642  {cab 2339   =/= wne 2517  (/)c0 3551

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-ne 2519  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-dif 3216  df-nul 3552

This theorem is referenced by:  setswith  4322  nnadjoin  4521  tfinnn  4535