Theorem soeq1 5610

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 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		poeq1 5592	. . 3 ⊢ (𝑅 = 𝑆 → (𝑅 Po 𝐴 ↔ 𝑆 Po 𝐴))
2		breq 5151	. . . . 5 ⊢ (𝑅 = 𝑆 → (𝑥𝑅𝑦 ↔ 𝑥𝑆𝑦))
3		biidd 262	. . . . 5 ⊢ (𝑅 = 𝑆 → (𝑥 = 𝑦 ↔ 𝑥 = 𝑦))
4		breq 5151	. . . . 5 ⊢ (𝑅 = 𝑆 → (𝑦𝑅𝑥 ↔ 𝑦𝑆𝑥))
5	2, 3, 4	3orbi123d 1436	. . . 4 ⊢ (𝑅 = 𝑆 → ((𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥) ↔ (𝑥𝑆𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑆𝑥)))
6	5	2ralbidv 3219	. . 3 ⊢ (𝑅 = 𝑆 → (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥) ↔ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥𝑆𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑆𝑥)))
7	1, 6	anbi12d 632	. 2 ⊢ (𝑅 = 𝑆 → ((𝑅 Po 𝐴 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥)) ↔ (𝑆 Po 𝐴 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥𝑆𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑆𝑥))))
8		df-so 5590	. 2 ⊢ (𝑅 Or 𝐴 ↔ (𝑅 Po 𝐴 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥)))
9		df-so 5590	. 2 ⊢ (𝑆 Or 𝐴 ↔ (𝑆 Po 𝐴 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥𝑆𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑆𝑥)))
10	7, 8, 9	3bitr4g 314	1 ⊢ (𝑅 = 𝑆 → (𝑅 Or 𝐴 ↔ 𝑆 Or 𝐴))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 397   ∨ w3o 1087   = wceq 1542  ∀wral 3062   class class class wbr 5149   Po wpo 5587   Or wor 5588

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-ext 2704

This theorem depends on definitions:  df-bi 206  df-an 398  df-or 847  df-3or 1089  df-ex 1783  df-cleq 2725  df-clel 2811  df-ral 3063  df-br 5150  df-po 5589  df-so 5590

This theorem is referenced by:  weeq1  5665  ltsopi  10883  cnso  16190  opsrtoslem2  21617  soeq12d  41780