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Theorem pzriprnglem3 21844
Description: Lemma 3 for pzriprng 21858: 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.
StepHypRef Expression
1 pzriprng.i . . 3 𝐼 = (ℤ × {0})
21eleq2i 2856 . 2 (𝑋 ∈ 𝐼 ↔ 𝑋 ∈ (ℤ × {0}))
3 elxp2 5682 . 2 (𝑋 ∈ (ℤ × {0}) ↔ ∃𝑥 ∈ ℤ ∃𝑦 ∈ {0}𝑋 = ⟨𝑥, 𝑦⟩)
4 0z 12744 . . . 4 0 ∈ ℤ
5 opeq2 4846 . . . . . 6 (𝑦 = 0 → ⟨𝑥, 𝑦⟩ = ⟨𝑥, 0⟩)
65eqeq2d 2774 . . . . 5 (𝑦 = 0 → (𝑋 = ⟨𝑥, 𝑦⟩ ↔ 𝑋 = ⟨𝑥, 0⟩))
76rexsng 4638 . . . 4 (0 ∈ ℤ → (∃𝑦 ∈ {0}𝑋 = ⟨𝑥, 𝑦⟩ ↔ 𝑋 = ⟨𝑥, 0⟩))
84, 7ax-mp 5 . . 3 (∃𝑦 ∈ {0}𝑋 = ⟨𝑥, 𝑦⟩ ↔ 𝑋 = ⟨𝑥, 0⟩)
98rexbii 3113 . 2 (∃𝑥 ∈ ℤ ∃𝑦 ∈ {0}𝑋 = ⟨𝑥, 𝑦⟩ ↔ ∃𝑥 ∈ ℤ 𝑋 = ⟨𝑥, 0⟩)
102, 3, 93bitri 300 1 (𝑋 ∈ 𝐼 ↔ ∃𝑥 ∈ ℤ 𝑋 = ⟨𝑥, 0⟩)
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   ↔ wb 209   = wceq 1570   ∈ wcel 2147  ∃wrex 3090  {csn 4588  ⟨cop 4840   × cxp 5656  (class class class)co 7451  0cc0 11239  ℤcz 12733   ×s cxps 17728  ℤringczring 21806
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735  ax-sep 5255  ax-pr 5398  ax-1cn 11297  ax-addrcl 11300  ax-rnegex 11310  ax-cnre 11312
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3or 1104  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ral 3081  df-rex 3091  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4841  df-uni 4873  df-br 5109  df-opab 5174  df-xp 5664  df-iota 6520  df-fv 6574  df-ov 7454  df-neg 11583  df-z 12734
This theorem is used by:  pzriprnglem4  21845  pzriprnglem5  21846  pzriprnglem6  21847  pzriprnglem8  21849
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