Theorem pzriprnglem3 21602

Description: Lemma 3 for pzriprng 21616: 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 2861	. 2 ⊢ (𝑋 ∈ 𝐼 ↔ 𝑋 ∈ (ℤ × {0}))
3		elxp2 5686	. 2 ⊢ (𝑋 ∈ (ℤ × {0}) ↔ ∃𝑥 ∈ ℤ ∃𝑦 ∈ {0}𝑋 = ⟨𝑥, 𝑦⟩)
4		0z 12602	. . . 4 ⊢ 0 ∈ ℤ
5		opeq2 4843	. . . . . 6 ⊢ (𝑦 = 0 → ⟨𝑥, 𝑦⟩ = ⟨𝑥, 0⟩)
6	5	eqeq2d 2780	. . . . 5 ⊢ (𝑦 = 0 → (𝑋 = ⟨𝑥, 𝑦⟩ ↔ 𝑋 = ⟨𝑥, 0⟩))
7	6	rexsng 4647	. . . 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 3118	. 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 1567   ∈ wcel 2149  ∃wrex 3095  {csn 4594  ⟨cop 4600   × cxp 5660  (class class class)co 7411  0cc0 11100  ℤcz 12591   ×_s cxps 17560  ℤ_ringczring 21565

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-ext 2741  ax-sep 5261  ax-pr 5405  ax-1cn 11158  ax-addrcl 11161  ax-rnegex 11171  ax-cnre 11173

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3or 1102  df-3an 1103  df-tru 1570  df-fal 1580  df-ex 1807  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-ral 3086  df-rex 3096  df-rab 3424  df-v 3465  df-dif 3916  df-un 3918  df-in 3920  df-ss 3930  df-nul 4295  df-if 4493  df-sn 4595  df-pr 4597  df-op 4601  df-uni 4877  df-br 5114  df-opab 5178  df-xp 5668  df-iota 6493  df-fv 6545  df-ov 7414  df-neg 11444  df-z 12592

This theorem is referenced by:  pzriprnglem4  21603  pzriprnglem5  21604  pzriprnglem6  21605  pzriprnglem8  21607