Theorem pzriprnglem3 21413

Description: Lemma 3 for pzriprng 21427: An element of 𝐼 is an ordered pair. (Contributed by AV, 18-Mar-2025.)

Hypotheses

Ref	Expression
pzriprng.r	⊢ 𝑅 = (ℤring ×s ℤring)
pzriprng.i	⊢ 𝐼 = (ℤ × {0})

Assertion

Ref	Expression
pzriprnglem3	⊢ (𝑋 ∈ 𝐼 ↔ ∃𝑥 ∈ ℤ 𝑋 = ⟨𝑥, 0⟩)

Distinct variable group:   𝑥,𝑋

Allowed substitution hints:   𝑅(𝑥)   𝐼(𝑥)

Proof of Theorem pzriprnglem3

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		pzriprng.i	. . 3 ⊢ 𝐼 = (ℤ × {0})
2	1	eleq2i 2817	. 2 ⊢ (𝑋 ∈ 𝐼 ↔ 𝑋 ∈ (ℤ × {0}))
3		elxp2 5696	. 2 ⊢ (𝑋 ∈ (ℤ × {0}) ↔ ∃𝑥 ∈ ℤ ∃𝑦 ∈ {0}𝑋 = ⟨𝑥, 𝑦⟩)
4		0z 12599	. . . 4 ⊢ 0 ∈ ℤ
5		opeq2 4870	. . . . . 6 ⊢ (𝑦 = 0 → ⟨𝑥, 𝑦⟩ = ⟨𝑥, 0⟩)
6	5	eqeq2d 2736	. . . . 5 ⊢ (𝑦 = 0 → (𝑋 = ⟨𝑥, 𝑦⟩ ↔ 𝑋 = ⟨𝑥, 0⟩))
7	6	rexsng 4674	. . . 4 ⊢ (0 ∈ ℤ → (∃𝑦 ∈ {0}𝑋 = ⟨𝑥, 𝑦⟩ ↔ 𝑋 = ⟨𝑥, 0⟩))
8	4, 7	ax-mp 5	. . 3 ⊢ (∃𝑦 ∈ {0}𝑋 = ⟨𝑥, 𝑦⟩ ↔ 𝑋 = ⟨𝑥, 0⟩)
9	8	rexbii 3084	. 2 ⊢ (∃𝑥 ∈ ℤ ∃𝑦 ∈ {0}𝑋 = ⟨𝑥, 𝑦⟩ ↔ ∃𝑥 ∈ ℤ 𝑋 = ⟨𝑥, 0⟩)
10	2, 3, 9	3bitri 296	1 ⊢ (𝑋 ∈ 𝐼 ↔ ∃𝑥 ∈ ℤ 𝑋 = ⟨𝑥, 0⟩)

Syntax hints:   ↔ wb 205   = wceq 1533   ∈ wcel 2098  ∃wrex 3060  {csn 4624  ⟨cop 4630   × cxp 5670  (class class class)co 7416  0cc0 11138  ℤcz 12588   ×_s cxps 17487  ℤ_ringczring 21376

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-ext 2696  ax-sep 5294  ax-nul 5301  ax-pr 5423  ax-1cn 11196  ax-addrcl 11199  ax-rnegex 11209  ax-cnre 11211

This theorem depends on definitions:  df-bi 206  df-an 395  df-or 846  df-3or 1085  df-3an 1086  df-tru 1536  df-fal 1546  df-ex 1774  df-sb 2060  df-clab 2703  df-cleq 2717  df-clel 2802  df-ral 3052  df-rex 3061  df-rab 3420  df-v 3465  df-dif 3942  df-un 3944  df-ss 3956  df-nul 4319  df-if 4525  df-sn 4625  df-pr 4627  df-op 4631  df-uni 4904  df-br 5144  df-opab 5206  df-xp 5678  df-iota 6495  df-fv 6551  df-ov 7419  df-neg 11477  df-z 12589

This theorem is referenced by:  pzriprnglem4  21414  pzriprnglem5  21415  pzriprnglem6  21416  pzriprnglem8  21418