Theorem pzriprnglem3 21400

Description: Lemma 3 for pzriprng 21414: 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 2821	. 2 ⊢ (𝑋 ∈ 𝐼 ↔ 𝑋 ∈ (ℤ × {0}))
3		elxp2 5665	. 2 ⊢ (𝑋 ∈ (ℤ × {0}) ↔ ∃𝑥 ∈ ℤ ∃𝑦 ∈ {0}𝑋 = ⟨𝑥, 𝑦⟩)
4		0z 12547	. . . 4 ⊢ 0 ∈ ℤ
5		opeq2 4841	. . . . . 6 ⊢ (𝑦 = 0 → ⟨𝑥, 𝑦⟩ = ⟨𝑥, 0⟩)
6	5	eqeq2d 2741	. . . . 5 ⊢ (𝑦 = 0 → (𝑋 = ⟨𝑥, 𝑦⟩ ↔ 𝑋 = ⟨𝑥, 0⟩))
7	6	rexsng 4643	. . . 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 3077	. 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 1540   ∈ wcel 2109  ∃wrex 3054  {csn 4592  ⟨cop 4598   × cxp 5639  (class class class)co 7390  0cc0 11075  ℤcz 12536   ×_s cxps 17476  ℤ_ringczring 21363

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2702  ax-sep 5254  ax-nul 5264  ax-pr 5390  ax-1cn 11133  ax-addrcl 11136  ax-rnegex 11146  ax-cnre 11148

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3or 1087  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2066  df-clab 2709  df-cleq 2722  df-clel 2804  df-ral 3046  df-rex 3055  df-rab 3409  df-v 3452  df-dif 3920  df-un 3922  df-ss 3934  df-nul 4300  df-if 4492  df-sn 4593  df-pr 4595  df-op 4599  df-uni 4875  df-br 5111  df-opab 5173  df-xp 5647  df-iota 6467  df-fv 6522  df-ov 7393  df-neg 11415  df-z 12537

This theorem is referenced by:  pzriprnglem4  21401  pzriprnglem5  21402  pzriprnglem6  21403  pzriprnglem8  21405