Theorem 0nelxpOLD 5144

Description: Obsolete proof of 0nelxp 5143 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 3203	. . . . . 6 ⊢ 𝑥 ∈ V
2		vex 3203	. . . . . 6 ⊢ 𝑦 ∈ V
3	1, 2	opnzi 4943	. . . . 5 ⊢ ⟨𝑥, 𝑦⟩ ≠ ∅
4		simpl 473	. . . . . . 7 ⊢ ((∅ = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)) → ∅ = ⟨𝑥, 𝑦⟩)
5	4	eqcomd 2628	. . . . . 6 ⊢ ((∅ = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)) → ⟨𝑥, 𝑦⟩ = ∅)
6	5	necon3ai 2819	. . . . 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 1731	. . 3 ⊢ ¬ ∃𝑦(∅ = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵))
9	8	nex 1731	. 2 ⊢ ¬ ∃𝑥∃𝑦(∅ = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵))
10		elxp 5131	. 2 ⊢ (∅ ∈ (𝐴 × 𝐵) ↔ ∃𝑥∃𝑦(∅ = ⟨𝑥, 𝑦⟩ ∧ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)))
11	9, 10	mtbir 313	1 ⊢ ¬ ∅ ∈ (𝐴 × 𝐵)

Syntax hints:  ¬ wn 3   ∧ wa 384   = wceq 1483  ∃wex 1704   ∈ wcel 1990   ≠ wne 2794  ∅c0 3915  ⟨cop 4183   × cxp 5112

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602  ax-sep 4781  ax-nul 4789  ax-pr 4906

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-3an 1039  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-ne 2795  df-v 3202  df-dif 3577  df-un 3579  df-in 3581  df-ss 3588  df-nul 3916  df-if 4087  df-sn 4178  df-pr 4180  df-op 4184  df-opab 4713  df-xp 5120

This theorem is referenced by: (None)