dmr0ss - Mathbox for BTernaryTau

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Theorem dmr0ss 36002

Description: The domain of 𝑅₀ is a subset of On × On. (Contributed by BTernaryTau, 22-Sep-2026.)

**Assertion**
Ref	Expression
dmr0ss	⊢ dom 𝑅₀ ⊆ (On × On)

Proof of Theorem dmr0ss Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		df-r0 35990	. . . 4 ⊢ 𝑅₀ = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ (On × On) ∧ 𝑦 ∈ (On × On)) ∧ (((1^st ‘𝑥) ∪ (2^nd ‘𝑥)) ∈ ((1^st ‘𝑦) ∪ (2^nd ‘𝑦)) ∨ (((1^st ‘𝑥) ∪ (2^nd ‘𝑥)) = ((1^st ‘𝑦) ∪ (2^nd ‘𝑦)) ∧ 𝑥LexOrd𝑦)))}
2		simpll 779	. . . . . 6 ⊢ (((𝑥 ∈ (On × On) ∧ 𝑦 ∈ (On × On)) ∧ (((1^st ‘𝑥) ∪ (2^nd ‘𝑥)) ∈ ((1^st ‘𝑦) ∪ (2^nd ‘𝑦)) ∨ (((1^st ‘𝑥) ∪ (2^nd ‘𝑥)) = ((1^st ‘𝑦) ∪ (2^nd ‘𝑦)) ∧ 𝑥LexOrd𝑦))) → 𝑥 ∈ (On × On))
3	2	pm4.71ri 570	. . . . 5 ⊢ (((𝑥 ∈ (On × On) ∧ 𝑦 ∈ (On × On)) ∧ (((1^st ‘𝑥) ∪ (2^nd ‘𝑥)) ∈ ((1^st ‘𝑦) ∪ (2^nd ‘𝑦)) ∨ (((1^st ‘𝑥) ∪ (2^nd ‘𝑥)) = ((1^st ‘𝑦) ∪ (2^nd ‘𝑦)) ∧ 𝑥LexOrd𝑦))) ↔ (𝑥 ∈ (On × On) ∧ ((𝑥 ∈ (On × On) ∧ 𝑦 ∈ (On × On)) ∧ (((1^st ‘𝑥) ∪ (2^nd ‘𝑥)) ∈ ((1^st ‘𝑦) ∪ (2^nd ‘𝑦)) ∨ (((1^st ‘𝑥) ∪ (2^nd ‘𝑥)) = ((1^st ‘𝑦) ∪ (2^nd ‘𝑦)) ∧ 𝑥LexOrd𝑦)))))
4	3	opabbii 5175	. . . 4 ⊢ {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ (On × On) ∧ 𝑦 ∈ (On × On)) ∧ (((1^st ‘𝑥) ∪ (2^nd ‘𝑥)) ∈ ((1^st ‘𝑦) ∪ (2^nd ‘𝑦)) ∨ (((1^st ‘𝑥) ∪ (2^nd ‘𝑥)) = ((1^st ‘𝑦) ∪ (2^nd ‘𝑦)) ∧ 𝑥LexOrd𝑦)))} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ (On × On) ∧ ((𝑥 ∈ (On × On) ∧ 𝑦 ∈ (On × On)) ∧ (((1^st ‘𝑥) ∪ (2^nd ‘𝑥)) ∈ ((1^st ‘𝑦) ∪ (2^nd ‘𝑦)) ∨ (((1^st ‘𝑥) ∪ (2^nd ‘𝑥)) = ((1^st ‘𝑦) ∪ (2^nd ‘𝑦)) ∧ 𝑥LexOrd𝑦))))}
5	1, 4	eqtri 2786	. . 3 ⊢ 𝑅₀ = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ (On × On) ∧ ((𝑥 ∈ (On × On) ∧ 𝑦 ∈ (On × On)) ∧ (((1^st ‘𝑥) ∪ (2^nd ‘𝑥)) ∈ ((1^st ‘𝑦) ∪ (2^nd ‘𝑦)) ∨ (((1^st ‘𝑥) ∪ (2^nd ‘𝑥)) = ((1^st ‘𝑦) ∪ (2^nd ‘𝑦)) ∧ 𝑥LexOrd𝑦))))}
6	5	dmeqi 5893	. 2 ⊢ dom 𝑅₀ = dom {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ (On × On) ∧ ((𝑥 ∈ (On × On) ∧ 𝑦 ∈ (On × On)) ∧ (((1^st ‘𝑥) ∪ (2^nd ‘𝑥)) ∈ ((1^st ‘𝑦) ∪ (2^nd ‘𝑦)) ∨ (((1^st ‘𝑥) ∪ (2^nd ‘𝑥)) = ((1^st ‘𝑦) ∪ (2^nd ‘𝑦)) ∧ 𝑥LexOrd𝑦))))}
7		dmopabss 5907	. 2 ⊢ dom {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ (On × On) ∧ ((𝑥 ∈ (On × On) ∧ 𝑦 ∈ (On × On)) ∧ (((1^st ‘𝑥) ∪ (2^nd ‘𝑥)) ∈ ((1^st ‘𝑦) ∪ (2^nd ‘𝑦)) ∨ (((1^st ‘𝑥) ∪ (2^nd ‘𝑥)) = ((1^st ‘𝑦) ∪ (2^nd ‘𝑦)) ∧ 𝑥LexOrd𝑦))))} ⊆ (On × On)
8	6, 7	eqsstri 3959	1 ⊢ dom 𝑅₀ ⊆ (On × On)

Colors of variables: wff setvar class

This proof depends on syntax axioms: ∧ wa 401 ∨ wo 861 = wceq 1570 ∈ wcel 2147 ⊆ wss 3900 ∪ cun 4152 class class class wbr 5106 {copab 5170 × cxp 5653 dom cdm 5655 Oncon0 6380 ‘cfv 6558 1^st c1st 8023 2^nd c2nd 8024 LexOrdclexo 35980 𝑅₀cr0 35981

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-10 2180 ax-11 2197 ax-12 2216 ax-ext 2735 ax-sep 5252 ax-pr 5395

This proof depends on definitions: df-bi 210 df-an 402 df-or 862 df-3an 1105 df-tru 1573 df-fal 1583 df-ex 1813 df-nf 1817 df-sb 2102 df-clab 2742 df-cleq 2755 df-clel 2839 df-nfc 2913 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 4839 df-br 5107 df-opab 5171 df-dm 5665 df-r0 35990

This theorem is used by: j0ltalephlem 36007

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