Theorem weeq2 4208

Description: Equality theorem for the well-ordering predicate. (Contributed by NM, 3-Apr-1994.)

Assertion

Ref	Expression
weeq2	⊢ (𝐴 = 𝐵 → (𝑅 We 𝐴 ↔ 𝑅 We 𝐵))

Proof of Theorem weeq2

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

Step	Hyp	Ref	Expression
1		freq2 4197	. . 3 ⊢ (𝐴 = 𝐵 → (𝑅 Fr 𝐴 ↔ 𝑅 Fr 𝐵))
2		raleq 2576	. . . . 5 ⊢ (𝐴 = 𝐵 → (∀𝑧 ∈ 𝐴 ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧) ↔ ∀𝑧 ∈ 𝐵 ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)))
3	2	raleqbi1dv 2584	. . . 4 ⊢ (𝐴 = 𝐵 → (∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧) ↔ ∀𝑦 ∈ 𝐵 ∀𝑧 ∈ 𝐵 ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)))
4	3	raleqbi1dv 2584	. . 3 ⊢ (𝐴 = 𝐵 → (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧) ↔ ∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 ∀𝑧 ∈ 𝐵 ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)))
5	1, 4	anbi12d 458	. 2 ⊢ (𝐴 = 𝐵 → ((𝑅 Fr 𝐴 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ↔ (𝑅 Fr 𝐵 ∧ ∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 ∀𝑧 ∈ 𝐵 ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧))))
6		df-wetr 4185	. 2 ⊢ (𝑅 We 𝐴 ↔ (𝑅 Fr 𝐴 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)))
7		df-wetr 4185	. 2 ⊢ (𝑅 We 𝐵 ↔ (𝑅 Fr 𝐵 ∧ ∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 ∀𝑧 ∈ 𝐵 ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)))
8	5, 6, 7	3bitr4g 222	1 ⊢ (𝐴 = 𝐵 → (𝑅 We 𝐴 ↔ 𝑅 We 𝐵))

Syntax hints:   → wi 4   ∧ wa 103   ↔ wb 104   = wceq 1296  ∀wral 2370   class class class wbr 3867   Fr wfr 4179   We wwe 4181

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-io 668  ax-5 1388  ax-7 1389  ax-gen 1390  ax-ie1 1434  ax-ie2 1435  ax-8 1447  ax-10 1448  ax-11 1449  ax-i12 1450  ax-bndl 1451  ax-4 1452  ax-17 1471  ax-i9 1475  ax-ial 1479  ax-i5r 1480  ax-ext 2077

This theorem depends on definitions:  df-bi 116  df-tru 1299  df-nf 1402  df-sb 1700  df-clab 2082  df-cleq 2088  df-clel 2091  df-nfc 2224  df-ral 2375  df-in 3019  df-ss 3026  df-frfor 4182  df-frind 4183  df-wetr 4185

This theorem is referenced by:  reg3exmid  4423