Theorem soeq1 4300

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 4284	. . 3 ⊢ (𝑅 = 𝑆 → (𝑅 Po 𝐴 ↔ 𝑆 Po 𝐴))
2		breq 3991	. . . . . 6 ⊢ (𝑅 = 𝑆 → (𝑥𝑅𝑦 ↔ 𝑥𝑆𝑦))
3		breq 3991	. . . . . . 7 ⊢ (𝑅 = 𝑆 → (𝑥𝑅𝑧 ↔ 𝑥𝑆𝑧))
4		breq 3991	. . . . . . 7 ⊢ (𝑅 = 𝑆 → (𝑧𝑅𝑦 ↔ 𝑧𝑆𝑦))
5	3, 4	orbi12d 788	. . . . . 6 ⊢ (𝑅 = 𝑆 → ((𝑥𝑅𝑧 ∨ 𝑧𝑅𝑦) ↔ (𝑥𝑆𝑧 ∨ 𝑧𝑆𝑦)))
6	2, 5	imbi12d 233	. . . . 5 ⊢ (𝑅 = 𝑆 → ((𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑧𝑅𝑦)) ↔ (𝑥𝑆𝑦 → (𝑥𝑆𝑧 ∨ 𝑧𝑆𝑦))))
7	6	2ralbidv 2494	. . . 4 ⊢ (𝑅 = 𝑆 → (∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑧𝑅𝑦)) ↔ ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑥𝑆𝑦 → (𝑥𝑆𝑧 ∨ 𝑧𝑆𝑦))))
8	7	ralbidv 2470	. . 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 470	. 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 4282	. 2 ⊢ (𝑅 Or 𝐴 ↔ (𝑅 Po 𝐴 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑧𝑅𝑦))))
11		df-iso 4282	. 2 ⊢ (𝑆 Or 𝐴 ↔ (𝑆 Po 𝐴 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑥𝑆𝑦 → (𝑥𝑆𝑧 ∨ 𝑧𝑆𝑦))))
12	9, 10, 11	3bitr4g 222	1 ⊢ (𝑅 = 𝑆 → (𝑅 Or 𝐴 ↔ 𝑆 Or 𝐴))

Syntax hints:   → wi 4   ∧ wa 103   ↔ wb 104   ∨ wo 703   = wceq 1348  ∀wral 2448   class class class wbr 3989   Po wpo 4279   Or wor 4280

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 609  ax-in2 610  ax-io 704  ax-5 1440  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-4 1503  ax-17 1519  ax-ial 1527  ax-ext 2152

This theorem depends on definitions:  df-bi 116  df-nf 1454  df-cleq 2163  df-clel 2166  df-ral 2453  df-br 3990  df-po 4281  df-iso 4282

This theorem is referenced by: (None)