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Theorem pzriprnglem3 21801
Description: Lemma 3 for pzriprng 21815: 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 2855 . 2 (𝑋 ∈ 𝐼 ↔ 𝑋 ∈ (ℤ × {0}))
3 elxp2 5678 . 2 (𝑋 ∈ (ℤ × {0}) ↔ ∃𝑥 ∈ ℤ ∃𝑦 ∈ {0}𝑋 = ⟨𝑥, 𝑦⟩)
4 0z 12714 . . . 4 0 ∈ ℤ
5 opeq2 4843 . . . . . 6 (𝑦 = 0 → ⟨𝑥, 𝑦⟩ = ⟨𝑥, 0⟩)
65eqeq2d 2774 . . . . 5 (𝑦 = 0 → (𝑋 = ⟨𝑥, 𝑦⟩ ↔ 𝑋 = ⟨𝑥, 0⟩))
76rexsng 4636 . . . 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 3112 . 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 1569   ∈ wcel 2146  ∃wrex 3089  {csn 4587  ⟨cop 4837   × cxp 5652  (class class class)co 7427  0cc0 11210  ℤcz 12703   ×s cxps 17690  ℤringczring 21764
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735  ax-sep 5251  ax-pr 5394  ax-1cn 11268  ax-addrcl 11271  ax-rnegex 11281  ax-cnre 11283
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3or 1104  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2838  df-ral 3080  df-rex 3090  df-rab 3416  df-v 3455  df-ss 3900  df-dif 4049  df-in 4093  df-un 4153  df-nul 4281  df-if 4485  df-sn 4588  df-pr 4590  df-op 4838  df-uni 4870  df-br 5106  df-opab 5170  df-xp 5660  df-iota 6500  df-fv 6554  df-ov 7430  df-neg 11554  df-z 12704
This theorem is used by:  pzriprnglem4  21802  pzriprnglem5  21803  pzriprnglem6  21804  pzriprnglem8  21806
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