Theorem ab0 3951

Description: The class of sets verifying a property is the empty class if and only if that property is a contradiction. See also abn0 3954 (from which it could be proved using as many essential proof steps but one fewer syntactic step, at the cost of depending on df-ne 2795). (Contributed by BJ, 19-Mar-2021.)

Assertion

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

Proof of Theorem ab0

Step	Hyp	Ref	Expression
1		nfab1 2766	. . 3 |- F/_ x { x | ph }
2	1	eq0f 3925	. 2 |- ( { x | ph } = (/) <-> A. x -. x e. { x | ph } )
3		abid 2610	. . . 4 |- ( x e. { x | ph } <-> ph )
4	3	notbii 310	. . 3 |- ( -. x e. { x | ph } <-> -. ph )
5	4	albii 1747	. 2 |- ( A. x -. x e. { x | ph } <-> A. x -. ph )
6	2, 5	bitri 264	1 |- ( { x | ph } = (/) <-> A. x -. ph )

Syntax hints:   -. wn 3   <-> wb 196   A.wal 1481   = wceq 1483   e. wcel 1990   {cab 2608   (/)c0 3915

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-v 3202  df-dif 3577  df-nul 3916

This theorem is referenced by:  dfnf5  3952  rab0  3955  rabeq0  3957  abf  3978