Metamath Proof Explorer


Theorem eq0rdv

Description: Deduction for equality to the empty set. (Contributed by NM, 11-Jul-2014) Avoid ax-8 , df-clel . (Revised by Gino Giotto, 6-Sep-2024)

Ref Expression
Hypothesis eq0rdv.1
|- ( ph -> -. x e. A )
Assertion eq0rdv
|- ( ph -> A = (/) )

Proof

Step Hyp Ref Expression
1 eq0rdv.1
 |-  ( ph -> -. x e. A )
2 1 alrimiv
 |-  ( ph -> A. x -. x e. A )
3 dfnul4
 |-  (/) = { y | F. }
4 3 eqeq2i
 |-  ( A = (/) <-> A = { y | F. } )
5 dfcleq
 |-  ( A = { y | F. } <-> A. x ( x e. A <-> x e. { y | F. } ) )
6 df-clab
 |-  ( x e. { y | F. } <-> [ x / y ] F. )
7 sbv
 |-  ( [ x / y ] F. <-> F. )
8 6 7 bitri
 |-  ( x e. { y | F. } <-> F. )
9 8 bibi2i
 |-  ( ( x e. A <-> x e. { y | F. } ) <-> ( x e. A <-> F. ) )
10 9 albii
 |-  ( A. x ( x e. A <-> x e. { y | F. } ) <-> A. x ( x e. A <-> F. ) )
11 nbfal
 |-  ( -. x e. A <-> ( x e. A <-> F. ) )
12 11 bicomi
 |-  ( ( x e. A <-> F. ) <-> -. x e. A )
13 12 albii
 |-  ( A. x ( x e. A <-> F. ) <-> A. x -. x e. A )
14 10 13 bitri
 |-  ( A. x ( x e. A <-> x e. { y | F. } ) <-> A. x -. x e. A )
15 4 5 14 3bitrri
 |-  ( A. x -. x e. A <-> A = (/) )
16 2 15 sylib
 |-  ( ph -> A = (/) )