Theorem nel0 3411

Description: From the general negation of membership in 𝐴, infer that 𝐴 is the empty set. (Contributed by BJ, 6-Oct-2018.)

Hypothesis

Ref	Expression
nel0.1	⊢ ¬ 𝑥 ∈ 𝐴

Assertion

Ref	Expression
nel0	⊢ 𝐴 = ∅

Distinct variable group:   𝑥,𝐴

Proof of Theorem nel0

Step	Hyp	Ref	Expression
1		eq0 3408	. 2 ⊢ (𝐴 = ∅ ↔ ∀𝑥 ¬ 𝑥 ∈ 𝐴)
2		nel0.1	. 2 ⊢ ¬ 𝑥 ∈ 𝐴
3	1, 2	mpgbir 1430	1 ⊢ 𝐴 = ∅

Syntax hints:  ¬ wn 3   = wceq 1332   ∈ wcel 2125  ∅c0 3390

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 604  ax-in2 605  ax-io 699  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1481  ax-10 1482  ax-11 1483  ax-i12 1484  ax-bndl 1486  ax-4 1487  ax-17 1503  ax-i9 1507  ax-ial 1511  ax-i5r 1512  ax-ext 2136

This theorem depends on definitions:  df-bi 116  df-tru 1335  df-nf 1438  df-sb 1740  df-clab 2141  df-cleq 2147  df-clel 2150  df-nfc 2285  df-v 2711  df-dif 3100  df-nul 3391

This theorem is referenced by: (None)