Theorem weeq2 2190

Description: Equality theorem for the well-ordering predicate.

Assertion

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

Proof of Theorem weeq2

Step	Hyp	Ref	Expression
1		freq2 2175	. . 3 ⊢ (A = B → (R Fr A ↔ R Fr B))
2		soeq2 2142	. . 3 ⊢ (A = B → (R Or A ↔ R Or B))
3	1, 2	anbi12d 476	. 2 ⊢ (A = B → ((R Fr A ∧ R Or A) ↔ (R Fr B ∧ R Or B)))
4		df-we 2186	. 2 ⊢ (R We A ↔ (R Fr A ∧ R Or A))
5		df-we 2186	. 2 ⊢ (R We B ↔ (R Fr B ∧ R Or B))
6	3, 4, 5	3bitr4g 428	1 ⊢ (A = B → (R We A ↔ R We B))

Syntax hints:   → wi 2   ↔ wb 127   ∧ wa 196   = wceq 1091   Or wor 2059   Fr wfr 2061   We wwe 2062

This theorem is referenced by:  ordeq 2206  hta 3619

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-3an 583  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-ral 1205  df-rex 1206  df-v 1349  df-dif 1489  df-un 1490  df-in 1491  df-ss 1492  df-nul 1708  df-sn 1811  df-pr 1812  df-op 1815  df-br 2063  df-po 2128  df-so 2138  df-fr 2169  df-we 2186

metamath.org