Theorem weeq2 4388

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 4377	. . 3 ⊢ (𝐴 = 𝐵 → (𝑅 Fr 𝐴 ↔ 𝑅 Fr 𝐵))
2		raleq 2690	. . . . 5 ⊢ (𝐴 = 𝐵 → (∀𝑧 ∈ 𝐴 ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧) ↔ ∀𝑧 ∈ 𝐵 ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)))
3	2	raleqbi1dv 2702	. . . 4 ⊢ (𝐴 = 𝐵 → (∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧) ↔ ∀𝑦 ∈ 𝐵 ∀𝑧 ∈ 𝐵 ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)))
4	3	raleqbi1dv 2702	. . 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 473	. 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 4365	. 2 ⊢ (𝑅 We 𝐴 ↔ (𝑅 Fr 𝐴 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)))
7		df-wetr 4365	. 2 ⊢ (𝑅 We 𝐵 ↔ (𝑅 Fr 𝐵 ∧ ∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 ∀𝑧 ∈ 𝐵 ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)))
8	5, 6, 7	3bitr4g 223	1 ⊢ (𝐴 = 𝐵 → (𝑅 We 𝐴 ↔ 𝑅 We 𝐵))

Syntax hints:   → wi 4   ∧ wa 104   ↔ wb 105   = wceq 1364  ∀wral 2472   class class class wbr 4029   Fr wfr 4359   We wwe 4361

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2175

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1472  df-sb 1774  df-clab 2180  df-cleq 2186  df-clel 2189  df-nfc 2325  df-ral 2477  df-in 3159  df-ss 3166  df-frfor 4362  df-frind 4363  df-wetr 4365

This theorem is referenced by:  reg3exmid  4612