Theorem pzriprnglem3 21633

Description: Lemma 3 for pzriprng 21647: 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 2855	. 2 ⊢ (𝑋 ∈ 𝐼 ↔ 𝑋 ∈ (ℤ × {0}))
3		elxp2 5685	. 2 ⊢ (𝑋 ∈ (ℤ × {0}) ↔ ∃𝑥 ∈ ℤ ∃𝑦 ∈ {0}𝑋 = ⟨𝑥, 𝑦⟩)
4		0z 12597	. . . 4 ⊢ 0 ∈ ℤ
5		opeq2 4839	. . . . . 6 ⊢ (𝑦 = 0 → ⟨𝑥, 𝑦⟩ = ⟨𝑥, 0⟩)
6	5	eqeq2d 2774	. . . . 5 ⊢ (𝑦 = 0 → (𝑋 = ⟨𝑥, 𝑦⟩ ↔ 𝑋 = ⟨𝑥, 0⟩))
7	6	rexsng 4642	. . . 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 3112	. 2 ⊢ (∃𝑥 ∈ ℤ ∃𝑦 ∈ {0}𝑋 = ⟨𝑥, 𝑦⟩ ↔ ∃𝑥 ∈ ℤ 𝑋 = ⟨𝑥, 0⟩)
10	2, 3, 9	3bitri 300	1 ⊢ (𝑋 ∈ 𝐼 ↔ ∃𝑥 ∈ ℤ 𝑋 = ⟨𝑥, 0⟩)

Syntax hints:   ↔ wb 209   = wceq 1570   ∈ wcel 2143  ∃wrex 3089  {csn 4589  ⟨cop 4595   × cxp 5659  (class class class)co 7410  0cc0 11095  ℤcz 12586   ×_s cxps 17555  ℤ_ringczring 21596

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-ext 2735  ax-sep 5257  ax-pr 5404  ax-1cn 11153  ax-addrcl 11156  ax-rnegex 11166  ax-cnre 11168

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3or 1104  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-ral 3080  df-rex 3090  df-rab 3417  df-v 3457  df-dif 3908  df-un 3910  df-in 3912  df-ss 3922  df-nul 4287  df-if 4488  df-sn 4590  df-pr 4592  df-op 4596  df-uni 4873  df-br 5110  df-opab 5174  df-xp 5667  df-iota 6492  df-fv 6544  df-ov 7413  df-neg 11439  df-z 12587

This theorem is referenced by:  pzriprnglem4  21634  pzriprnglem5  21635  pzriprnglem6  21636  pzriprnglem8  21638