Theorem weeq1 4194

Description: Equality theorem for the well-ordering predicate. (Contributed by NM, 9-Mar-1997.)

Assertion

Ref	Expression
weeq1	⊢ (𝑅 = 𝑆 → (𝑅 We 𝐴 ↔ 𝑆 We 𝐴))

Proof of Theorem weeq1

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

Step	Hyp	Ref	Expression
1		freq1 4182	. . 3 ⊢ (𝑅 = 𝑆 → (𝑅 Fr 𝐴 ↔ 𝑆 Fr 𝐴))
2		breq 3855	. . . . . . . 8 ⊢ (𝑅 = 𝑆 → (𝑥𝑅𝑦 ↔ 𝑥𝑆𝑦))
3		breq 3855	. . . . . . . 8 ⊢ (𝑅 = 𝑆 → (𝑦𝑅𝑧 ↔ 𝑦𝑆𝑧))
4	2, 3	anbi12d 458	. . . . . . 7 ⊢ (𝑅 = 𝑆 → ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) ↔ (𝑥𝑆𝑦 ∧ 𝑦𝑆𝑧)))
5		breq 3855	. . . . . . 7 ⊢ (𝑅 = 𝑆 → (𝑥𝑅𝑧 ↔ 𝑥𝑆𝑧))
6	4, 5	imbi12d 233	. . . . . 6 ⊢ (𝑅 = 𝑆 → (((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧) ↔ ((𝑥𝑆𝑦 ∧ 𝑦𝑆𝑧) → 𝑥𝑆𝑧)))
7	6	ralbidv 2381	. . . . 5 ⊢ (𝑅 = 𝑆 → (∀𝑧 ∈ 𝐴 ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧) ↔ ∀𝑧 ∈ 𝐴 ((𝑥𝑆𝑦 ∧ 𝑦𝑆𝑧) → 𝑥𝑆𝑧)))
8	7	ralbidv 2381	. . . 4 ⊢ (𝑅 = 𝑆 → (∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧) ↔ ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 ((𝑥𝑆𝑦 ∧ 𝑦𝑆𝑧) → 𝑥𝑆𝑧)))
9	8	ralbidv 2381	. . 3 ⊢ (𝑅 = 𝑆 → (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧) ↔ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 ((𝑥𝑆𝑦 ∧ 𝑦𝑆𝑧) → 𝑥𝑆𝑧)))
10	1, 9	anbi12d 458	. 2 ⊢ (𝑅 = 𝑆 → ((𝑅 Fr 𝐴 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ↔ (𝑆 Fr 𝐴 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 ((𝑥𝑆𝑦 ∧ 𝑦𝑆𝑧) → 𝑥𝑆𝑧))))
11		df-wetr 4172	. 2 ⊢ (𝑅 We 𝐴 ↔ (𝑅 Fr 𝐴 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)))
12		df-wetr 4172	. 2 ⊢ (𝑆 We 𝐴 ↔ (𝑆 Fr 𝐴 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 ((𝑥𝑆𝑦 ∧ 𝑦𝑆𝑧) → 𝑥𝑆𝑧)))
13	10, 11, 12	3bitr4g 222	1 ⊢ (𝑅 = 𝑆 → (𝑅 We 𝐴 ↔ 𝑆 We 𝐴))

Syntax hints:   → wi 4   ∧ wa 103   ↔ wb 104   = wceq 1290  ∀wral 2360   class class class wbr 3853   Fr wfr 4166   We wwe 4168

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1382  ax-gen 1384  ax-ie1 1428  ax-ie2 1429  ax-4 1446  ax-17 1465  ax-ial 1473  ax-ext 2071

This theorem depends on definitions:  df-bi 116  df-nf 1396  df-cleq 2082  df-clel 2085  df-ral 2365  df-br 3854  df-frfor 4169  df-frind 4170  df-wetr 4172

This theorem is referenced by: (None)