Theorem eq0rdv 3585

Description: Deduction rule for equality to the empty set. (Contributed by NM, 11-Jul-2014.)

Hypothesis

Ref	Expression
eq0rdv.1	|- (ph -> -. x e. A)

Assertion

Ref	Expression
eq0rdv	|- (ph -> A = (/))

Distinct variable groups:   x,A   ph,x

Proof of Theorem eq0rdv

Step	Hyp	Ref	Expression
1		eq0rdv.1	. . . 4 |- (ph -> -. x e. A)
2	1	pm2.21d 98	. . 3 |- (ph -> (x e. A -> x e. (/)))
3	2	ssrdv 3278	. 2 |- (ph -> A C_ (/))
4		ss0 3581	. 2 |- (A C_ (/) -> A = (/))
5	3, 4	syl 15	1 |- (ph -> A = (/))

Syntax hints:  -. wn 3   -> wi 4   = wceq 1642   e. wcel 1710   C_ wss 3257  (/)c0 3550

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 2478  df-v 2861  df-nin 3211  df-compl 3212  df-in 3213  df-dif 3215  df-ss 3259  df-nul 3551

This theorem is referenced by:  tz6.12-2  5346  nchoicelem3  6291