Theorem pzriprnglem3 21453

Description: Lemma 3 for pzriprng 21467: 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 2829	. 2 ⊢ (𝑋 ∈ 𝐼 ↔ 𝑋 ∈ (ℤ × {0}))
3		elxp2 5656	. 2 ⊢ (𝑋 ∈ (ℤ × {0}) ↔ ∃𝑥 ∈ ℤ ∃𝑦 ∈ {0}𝑋 = ⟨𝑥, 𝑦⟩)
4		0z 12511	. . . 4 ⊢ 0 ∈ ℤ
5		opeq2 4832	. . . . . 6 ⊢ (𝑦 = 0 → ⟨𝑥, 𝑦⟩ = ⟨𝑥, 0⟩)
6	5	eqeq2d 2748	. . . . 5 ⊢ (𝑦 = 0 → (𝑋 = ⟨𝑥, 𝑦⟩ ↔ 𝑋 = ⟨𝑥, 0⟩))
7	6	rexsng 4635	. . . 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 3085	. 2 ⊢ (∃𝑥 ∈ ℤ ∃𝑦 ∈ {0}𝑋 = ⟨𝑥, 𝑦⟩ ↔ ∃𝑥 ∈ ℤ 𝑋 = ⟨𝑥, 0⟩)
10	2, 3, 9	3bitri 297	1 ⊢ (𝑋 ∈ 𝐼 ↔ ∃𝑥 ∈ ℤ 𝑋 = ⟨𝑥, 0⟩)

Syntax hints:   ↔ wb 206   = wceq 1542   ∈ wcel 2114  ∃wrex 3062  {csn 4582  ⟨cop 4588   × cxp 5630  (class class class)co 7368  0cc0 11038  ℤcz 12500   ×_s cxps 17439  ℤ_ringczring 21416

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709  ax-sep 5243  ax-pr 5379  ax-1cn 11096  ax-addrcl 11099  ax-rnegex 11109  ax-cnre 11111

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3or 1088  df-3an 1089  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-ral 3053  df-rex 3063  df-rab 3402  df-v 3444  df-dif 3906  df-un 3908  df-in 3910  df-ss 3920  df-nul 4288  df-if 4482  df-sn 4583  df-pr 4585  df-op 4589  df-uni 4866  df-br 5101  df-opab 5163  df-xp 5638  df-iota 6456  df-fv 6508  df-ov 7371  df-neg 11379  df-z 12501

This theorem is referenced by:  pzriprnglem4  21454  pzriprnglem5  21455  pzriprnglem6  21456  pzriprnglem8  21458