Theorem weeq1 4341

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 4329	. . 3 ⊢ (𝑅 = 𝑆 → (𝑅 Fr 𝐴 ↔ 𝑆 Fr 𝐴))
2		breq 3991	. . . . . . . 8 ⊢ (𝑅 = 𝑆 → (𝑥𝑅𝑦 ↔ 𝑥𝑆𝑦))
3		breq 3991	. . . . . . . 8 ⊢ (𝑅 = 𝑆 → (𝑦𝑅𝑧 ↔ 𝑦𝑆𝑧))
4	2, 3	anbi12d 470	. . . . . . 7 ⊢ (𝑅 = 𝑆 → ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) ↔ (𝑥𝑆𝑦 ∧ 𝑦𝑆𝑧)))
5		breq 3991	. . . . . . 7 ⊢ (𝑅 = 𝑆 → (𝑥𝑅𝑧 ↔ 𝑥𝑆𝑧))
6	4, 5	imbi12d 233	. . . . . 6 ⊢ (𝑅 = 𝑆 → (((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧) ↔ ((𝑥𝑆𝑦 ∧ 𝑦𝑆𝑧) → 𝑥𝑆𝑧)))
7	6	ralbidv 2470	. . . . 5 ⊢ (𝑅 = 𝑆 → (∀𝑧 ∈ 𝐴 ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧) ↔ ∀𝑧 ∈ 𝐴 ((𝑥𝑆𝑦 ∧ 𝑦𝑆𝑧) → 𝑥𝑆𝑧)))
8	7	ralbidv 2470	. . . 4 ⊢ (𝑅 = 𝑆 → (∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧) ↔ ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 ((𝑥𝑆𝑦 ∧ 𝑦𝑆𝑧) → 𝑥𝑆𝑧)))
9	8	ralbidv 2470	. . 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 470	. 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 4319	. 2 ⊢ (𝑅 We 𝐴 ↔ (𝑅 Fr 𝐴 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)))
12		df-wetr 4319	. 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 1348  ∀wral 2448   class class class wbr 3989   Fr wfr 4313   We wwe 4315

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-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-frfor 4316  df-frind 4317  df-wetr 4319

This theorem is referenced by: (None)