Theorem soeq1 5618

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 5600	. . 3 ⊢ (𝑅 = 𝑆 → (𝑅 Po 𝐴 ↔ 𝑆 Po 𝐴))
2		breq 5150	. . . . 5 ⊢ (𝑅 = 𝑆 → (𝑥𝑅𝑦 ↔ 𝑥𝑆𝑦))
3		biidd 262	. . . . 5 ⊢ (𝑅 = 𝑆 → (𝑥 = 𝑦 ↔ 𝑥 = 𝑦))
4		breq 5150	. . . . 5 ⊢ (𝑅 = 𝑆 → (𝑦𝑅𝑥 ↔ 𝑦𝑆𝑥))
5	2, 3, 4	3orbi123d 1434	. . . 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 5598	. 2 ⊢ (𝑅 Or 𝐴 ↔ (𝑅 Po 𝐴 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥)))
9		df-so 5598	. 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 206   ∧ wa 395   ∨ w3o 1085   = wceq 1537  ∀wral 3059   class class class wbr 5148   Po wpo 5595   Or wor 5596

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3or 1087  df-ex 1777  df-cleq 2727  df-clel 2814  df-ral 3060  df-br 5149  df-po 5597  df-so 5598

This theorem is referenced by:  soeq12d  5620  weeq1  5676  ltsopi  10926  cnso  16280