Theorem exexneq 5433

Description: There exist two different sets. (Contributed by NM, 7-Nov-2006.) Avoid ax-13 2369. (Revised by BJ, 31-May-2019.) Avoid ax-8 2106. (Revised by SN, 21-Sep-2023.) Avoid ax-12 2169. (Revised by Rohan Ridenour, 9-Oct-2024.) Use ax-pr 5426 instead of ax-pow 5362. (Revised by BTernaryTau, 3-Dec-2024.) Extract this result from the proof of dtru 5435. (Revised by BJ, 2-Jan-2025.)

Assertion

Ref	Expression
exexneq	⊢ ∃𝑥∃𝑦 ¬ 𝑥 = 𝑦

Distinct variable group:   𝑥,𝑦

Proof of Theorem exexneq

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		exel 5432	. . 3 ⊢ ∃𝑥∃𝑧 𝑧 ∈ 𝑥
2		ax-nul 5305	. . 3 ⊢ ∃𝑦∀𝑧 ¬ 𝑧 ∈ 𝑦
3		exdistrv 1957	. . 3 ⊢ (∃𝑥∃𝑦(∃𝑧 𝑧 ∈ 𝑥 ∧ ∀𝑧 ¬ 𝑧 ∈ 𝑦) ↔ (∃𝑥∃𝑧 𝑧 ∈ 𝑥 ∧ ∃𝑦∀𝑧 ¬ 𝑧 ∈ 𝑦))
4	1, 2, 3	mpbir2an 707	. 2 ⊢ ∃𝑥∃𝑦(∃𝑧 𝑧 ∈ 𝑥 ∧ ∀𝑧 ¬ 𝑧 ∈ 𝑦)
5		ax9v1 2116	. . . . . . . 8 ⊢ (𝑥 = 𝑦 → (𝑧 ∈ 𝑥 → 𝑧 ∈ 𝑦))
6	5	eximdv 1918	. . . . . . 7 ⊢ (𝑥 = 𝑦 → (∃𝑧 𝑧 ∈ 𝑥 → ∃𝑧 𝑧 ∈ 𝑦))
7		df-ex 1780	. . . . . . 7 ⊢ (∃𝑧 𝑧 ∈ 𝑦 ↔ ¬ ∀𝑧 ¬ 𝑧 ∈ 𝑦)
8	6, 7	imbitrdi 250	. . . . . 6 ⊢ (𝑥 = 𝑦 → (∃𝑧 𝑧 ∈ 𝑥 → ¬ ∀𝑧 ¬ 𝑧 ∈ 𝑦))
9	8	com12 32	. . . . 5 ⊢ (∃𝑧 𝑧 ∈ 𝑥 → (𝑥 = 𝑦 → ¬ ∀𝑧 ¬ 𝑧 ∈ 𝑦))
10	9	con2d 134	. . . 4 ⊢ (∃𝑧 𝑧 ∈ 𝑥 → (∀𝑧 ¬ 𝑧 ∈ 𝑦 → ¬ 𝑥 = 𝑦))
11	10	imp 405	. . 3 ⊢ ((∃𝑧 𝑧 ∈ 𝑥 ∧ ∀𝑧 ¬ 𝑧 ∈ 𝑦) → ¬ 𝑥 = 𝑦)
12	11	2eximi 1836	. 2 ⊢ (∃𝑥∃𝑦(∃𝑧 𝑧 ∈ 𝑥 ∧ ∀𝑧 ¬ 𝑧 ∈ 𝑦) → ∃𝑥∃𝑦 ¬ 𝑥 = 𝑦)
13	4, 12	ax-mp 5	1 ⊢ ∃𝑥∃𝑦 ¬ 𝑥 = 𝑦

Syntax hints:  ¬ wn 3   ∧ wa 394  ∀wal 1537  ∃wex 1779

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1911  ax-6 1969  ax-9 2114  ax-nul 5305  ax-pr 5426

This theorem depends on definitions:  df-bi 206  df-an 395  df-or 844  df-ex 1780

This theorem is referenced by:  exneq  5434