Theorem eq0rdv

Description: Deduction for equality to the empty set. (Contributed by NM, 11-Jul-2014) Avoid ax-8 , df-clel . (Revised by GG, 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		biidd	|- ( y = x -> ( F. <-> F. ) )
4	3	eqabbw	|- ( A = { y | F. } <-> A. x ( x e. A <-> F. ) )
5		dfnul4	|- (/) = { y | F. }
6	5	eqeq2i	|- ( A = (/) <-> A = { y | F. } )
7		nbfal	|- ( -. x e. A <-> ( x e. A <-> F. ) )
8	7	albii	|- ( A. x -. x e. A <-> A. x ( x e. A <-> F. ) )
9	4 6 8	3bitr4ri	|- ( A. x -. x e. A <-> A = (/) )
10	2 9	sylib	|- ( ph -> A = (/) )