Theorem 0nelxpOLD 5068

Description: Obsolete proof of 0nelxp 5067 as of 13-Aug-2021. (Contributed by NM, 2-May-1996.) (Revised by Mario Carneiro, 26-Apr-2015.) (New usage is discouraged.) (Proof modification is discouraged.)

Assertion

Ref	Expression
0nelxpOLD	⊢ ¬ ∅ ∈ (𝐴 × 𝐵)

Proof of Theorem 0nelxpOLD

Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		vex 3176	. . . . . 6 ⊢ 𝑥 ∈ V
2		vex 3176	. . . . . 6 ⊢ 𝑦 ∈ V
3	1, 2	opnzi 4869	. . . . 5 ⊢ ⟨𝑥, 𝑦⟩ ≠ ∅
4		simpl 472	. . . . . . 7 ⊢ ((∅ = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)) → ∅ = ⟨𝑥, 𝑦⟩)
5	4	eqcomd 2616	. . . . . 6 ⊢ ((∅ = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)) → ⟨𝑥, 𝑦⟩ = ∅)
6	5	necon3ai 2807	. . . . 5 ⊢ (⟨𝑥, 𝑦⟩ ≠ ∅ → ¬ (∅ = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)))
7	3, 6	ax-mp 5	. . . 4 ⊢ ¬ (∅ = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵))
8	7	nex 1722	. . 3 ⊢ ¬ ∃𝑦(∅ = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵))
9	8	nex 1722	. 2 ⊢ ¬ ∃𝑥∃𝑦(∅ = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵))
10		elxp 5055	. 2 ⊢ (∅ ∈ (𝐴 × 𝐵) ↔ ∃𝑥∃𝑦(∅ = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)))
11	9, 10	mtbir 312	1 ⊢ ¬ ∅ ∈ (𝐴 × 𝐵)

Syntax hints:  ¬ wn 3   ∧ wa 383   = wceq 1475  ∃wex 1695   ∈ wcel 1977   ≠ wne 2780  ∅c0 3874  ⟨cop 4131   × cxp 5036

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-9 1986  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590  ax-sep 4709  ax-nul 4717  ax-pr 4833

This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-3an 1033  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-ne 2782  df-v 3175  df-dif 3543  df-un 3545  df-in 3547  df-ss 3554  df-nul 3875  df-if 4037  df-sn 4126  df-pr 4128  df-op 4132  df-opab 4644  df-xp 5044

This theorem is referenced by: (None)