Theorem soeq1 4070

Description: Equality theorem for the strict ordering predicate. (Contributed by NM, 16-Mar-1997.)

Assertion

Ref	Expression
soeq1	⊢ (𝑅 = 𝑆 → (𝑅 Or 𝐴 ↔ 𝑆 Or 𝐴))

Proof of Theorem soeq1

Dummy variables 𝑥 𝑦 𝑧 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		poeq1 4054	. . 3 ⊢ (𝑅 = 𝑆 → (𝑅 Po 𝐴 ↔ 𝑆 Po 𝐴))
2		breq 3787	. . . . . 6 ⊢ (𝑅 = 𝑆 → (𝑥𝑅𝑦 ↔ 𝑥𝑆𝑦))
3		breq 3787	. . . . . . 7 ⊢ (𝑅 = 𝑆 → (𝑥𝑅𝑧 ↔ 𝑥𝑆𝑧))
4		breq 3787	. . . . . . 7 ⊢ (𝑅 = 𝑆 → (𝑧𝑅𝑦 ↔ 𝑧𝑆𝑦))
5	3, 4	orbi12d 739	. . . . . 6 ⊢ (𝑅 = 𝑆 → ((𝑥𝑅𝑧 ∨ 𝑧𝑅𝑦) ↔ (𝑥𝑆𝑧 ∨ 𝑧𝑆𝑦)))
6	2, 5	imbi12d 232	. . . . 5 ⊢ (𝑅 = 𝑆 → ((𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑧𝑅𝑦)) ↔ (𝑥𝑆𝑦 → (𝑥𝑆𝑧 ∨ 𝑧𝑆𝑦))))
7	6	2ralbidv 2390	. . . 4 ⊢ (𝑅 = 𝑆 → (∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑧𝑅𝑦)) ↔ ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑥𝑆𝑦 → (𝑥𝑆𝑧 ∨ 𝑧𝑆𝑦))))
8	7	ralbidv 2368	. . 3 ⊢ (𝑅 = 𝑆 → (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑧𝑅𝑦)) ↔ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑥𝑆𝑦 → (𝑥𝑆𝑧 ∨ 𝑧𝑆𝑦))))
9	1, 8	anbi12d 456	. 2 ⊢ (𝑅 = 𝑆 → ((𝑅 Po 𝐴 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑧𝑅𝑦))) ↔ (𝑆 Po 𝐴 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑥𝑆𝑦 → (𝑥𝑆𝑧 ∨ 𝑧𝑆𝑦)))))
10		df-iso 4052	. 2 ⊢ (𝑅 Or 𝐴 ↔ (𝑅 Po 𝐴 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑧𝑅𝑦))))
11		df-iso 4052	. 2 ⊢ (𝑆 Or 𝐴 ↔ (𝑆 Po 𝐴 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑥𝑆𝑦 → (𝑥𝑆𝑧 ∨ 𝑧𝑆𝑦))))
12	9, 10, 11	3bitr4g 221	1 ⊢ (𝑅 = 𝑆 → (𝑅 Or 𝐴 ↔ 𝑆 Or 𝐴))

Syntax hints:   → wi 4   ∧ wa 102   ↔ wb 103   ∨ wo 661   = wceq 1284  ∀wral 2348   class class class wbr 3785   Po wpo 4049   Or wor 4050

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in1 576  ax-in2 577  ax-io 662  ax-5 1376  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-4 1440  ax-17 1459  ax-ial 1467  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-nf 1390  df-cleq 2074  df-clel 2077  df-ral 2353  df-br 3786  df-po 4051  df-iso 4052

This theorem is referenced by: (None)