Theorem weeq1 4093

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 4081	. . 3 ⊢ (𝑅 = 𝑆 → (𝑅 Fr 𝐴 ↔ 𝑆 Fr 𝐴))
2		breq 3766	. . . . . . . 8 ⊢ (𝑅 = 𝑆 → (𝑥𝑅𝑦 ↔ 𝑥𝑆𝑦))
3		breq 3766	. . . . . . . 8 ⊢ (𝑅 = 𝑆 → (𝑦𝑅𝑧 ↔ 𝑦𝑆𝑧))
4	2, 3	anbi12d 442	. . . . . . 7 ⊢ (𝑅 = 𝑆 → ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) ↔ (𝑥𝑆𝑦 ∧ 𝑦𝑆𝑧)))
5		breq 3766	. . . . . . 7 ⊢ (𝑅 = 𝑆 → (𝑥𝑅𝑧 ↔ 𝑥𝑆𝑧))
6	4, 5	imbi12d 223	. . . . . 6 ⊢ (𝑅 = 𝑆 → (((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧) ↔ ((𝑥𝑆𝑦 ∧ 𝑦𝑆𝑧) → 𝑥𝑆𝑧)))
7	6	ralbidv 2326	. . . . 5 ⊢ (𝑅 = 𝑆 → (∀𝑧 ∈ 𝐴 ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧) ↔ ∀𝑧 ∈ 𝐴 ((𝑥𝑆𝑦 ∧ 𝑦𝑆𝑧) → 𝑥𝑆𝑧)))
8	7	ralbidv 2326	. . . 4 ⊢ (𝑅 = 𝑆 → (∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧) ↔ ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 ((𝑥𝑆𝑦 ∧ 𝑦𝑆𝑧) → 𝑥𝑆𝑧)))
9	8	ralbidv 2326	. . 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 442	. 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 4071	. 2 ⊢ (𝑅 We 𝐴 ↔ (𝑅 Fr 𝐴 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)))
12		df-wetr 4071	. 2 ⊢ (𝑆 We 𝐴 ↔ (𝑆 Fr 𝐴 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 ((𝑥𝑆𝑦 ∧ 𝑦𝑆𝑧) → 𝑥𝑆𝑧)))
13	10, 11, 12	3bitr4g 212	1 ⊢ (𝑅 = 𝑆 → (𝑅 We 𝐴 ↔ 𝑆 We 𝐴))

Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98   = wceq 1243  ∀wral 2306   class class class wbr 3764   Fr wfr 4065   We wwe 4067

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1336  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-4 1400  ax-17 1419  ax-ial 1427  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-nf 1350  df-cleq 2033  df-clel 2036  df-ral 2311  df-br 3765  df-frfor 4068  df-frind 4069  df-wetr 4071

This theorem is referenced by: (None)