Theorem notzfaus 5325

Description: In the Separation Scheme zfauscl 5253, we require that 𝑦 not occur in 𝜑 (which can be generalized to "not be free in"). Here we show special cases of 𝐴 and 𝜑 that result in a contradiction if that requirement is not met. (Contributed by NM, 8-Feb-2006.) (Proof shortened by BJ, 18-Nov-2023.)

Hypotheses

Ref	Expression
notzfaus.1	⊢ 𝐴 = {∅}
notzfaus.2	⊢ (𝜑 ↔ ¬ 𝑥 ∈ 𝑦)

Assertion

Ref	Expression
notzfaus	⊢ ¬ ∃𝑦∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))

Distinct variable group:   𝑥,𝐴

Allowed substitution hints:   𝜑(𝑥,𝑦)   𝐴(𝑦)

Proof of Theorem notzfaus

Step	Hyp	Ref	Expression
1		notzfaus.1	. . . . . 6 ⊢ 𝐴 = {∅}
2		0ex 5262	. . . . . . 7 ⊢ ∅ ∈ V
3	2	snnz 4738	. . . . . 6 ⊢ {∅} ≠ ∅
4	1, 3	eqnetri 3030	. . . . 5 ⊢ 𝐴 ≠ ∅
5		n0 4308	. . . . 5 ⊢ (𝐴 ≠ ∅ ↔ ∃𝑥 𝑥 ∈ 𝐴)
6	4, 5	mpbi 233	. . . 4 ⊢ ∃𝑥 𝑥 ∈ 𝐴
7		pm5.19 390	. . . . 5 ⊢ ¬ (𝑥 ∈ 𝑦 ↔ ¬ 𝑥 ∈ 𝑦)
8		ibar 537	. . . . . . 7 ⊢ (𝑥 ∈ 𝐴 → (𝜑 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑)))
9		notzfaus.2	. . . . . . 7 ⊢ (𝜑 ↔ ¬ 𝑥 ∈ 𝑦)
10	8, 9	bitr3di 289	. . . . . 6 ⊢ (𝑥 ∈ 𝐴 → ((𝑥 ∈ 𝐴 ∧ 𝜑) ↔ ¬ 𝑥 ∈ 𝑦))
11	10	bibi2d 345	. . . . 5 ⊢ (𝑥 ∈ 𝐴 → ((𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑)) ↔ (𝑥 ∈ 𝑦 ↔ ¬ 𝑥 ∈ 𝑦)))
12	7, 11	mtbiri 330	. . . 4 ⊢ (𝑥 ∈ 𝐴 → ¬ (𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑)))
13	6, 12	eximii 1860	. . 3 ⊢ ∃𝑥 ¬ (𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))
14		exnal 1850	. . 3 ⊢ (∃𝑥 ¬ (𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑)) ↔ ¬ ∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑)))
15	13, 14	mpbi 233	. 2 ⊢ ¬ ∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))
16	15	nex 1823	1 ⊢ ¬ ∃𝑦∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))

Syntax hints:  ¬ wn 3   ↔ wb 209   ∧ wa 400  ∀wal 1561   = wceq 1563  ∃wex 1802   ∈ wcel 2145   ≠ wne 2960  ∅c0 4288  {csn 4585

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1818  ax-4 1832  ax-5 1933  ax-6 1990  ax-7 2031  ax-8 2147  ax-9 2155  ax-ext 2737  ax-nul 5261

This theorem depends on definitions:  df-bi 210  df-an 401  df-tru 1566  df-fal 1576  df-ex 1803  df-sb 2094  df-clab 2744  df-cleq 2757  df-clel 2840  df-ne 2961  df-v 3459  df-dif 3910  df-nul 4289  df-sn 4586

This theorem is referenced by: (None)