Theorem weeq1 2189

Description: Equality theorem for the well-ordering predicate.

Assertion

Ref	Expression
weeq1	⊢ (R = S → (R We A ↔ S We A))

Proof of Theorem weeq1

Step	Hyp	Ref	Expression
1		freq1 2174	. . 3 ⊢ (R = S → (R Fr A ↔ S Fr A))
2		soeq1 2141	. . 3 ⊢ (R = S → (R Or A ↔ S Or A))
3	1, 2	anbi12d 476	. 2 ⊢ (R = S → ((R Fr A ∧ R Or A) ↔ (S Fr A ∧ S Or A)))
4		df-we 2186	. 2 ⊢ (R We A ↔ (R Fr A ∧ R Or A))
5		df-we 2186	. 2 ⊢ (S We A ↔ (S Fr A ∧ S Or A))
6	3, 4, 5	3bitr4g 428	1 ⊢ (R = S → (R We A ↔ S We A))

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

This theorem is referenced by:  weth 3602

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-gen 677  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-3or 582  df-ex 679  df-cleq 1097  df-clel 1099  df-ral 1205  df-rex 1206  df-br 2063  df-po 2128  df-so 2138  df-fr 2169  df-we 2186

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