Theorem eq0rdv 3586

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

Hypothesis

Ref	Expression
eq0rdv.1	⊢ (φ → ¬ x ∈ A)

Assertion

Ref	Expression
eq0rdv	⊢ (φ → A = ∅)

Distinct variable groups:   x,A   φ,x

Proof of Theorem eq0rdv

Step	Hyp	Ref	Expression
1		eq0rdv.1	. . . 4 ⊢ (φ → ¬ x ∈ A)
2	1	pm2.21d 98	. . 3 ⊢ (φ → (x ∈ A → x ∈ ∅))
3	2	ssrdv 3279	. 2 ⊢ (φ → A ⊆ ∅)
4		ss0 3582	. 2 ⊢ (A ⊆ ∅ → A = ∅)
5	3, 4	syl 15	1 ⊢ (φ → A = ∅)

Syntax hints:  ¬ wn 3   → wi 4   = wceq 1642   ∈ wcel 1710   ⊆ wss 3258  ∅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-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-dif 3216  df-ss 3260  df-nul 3552

This theorem is referenced by:  tz6.12-2  5347  nchoicelem3  6292