Theorem freq1 4182

Description: Equality theorem for the well-founded predicate. (Contributed by NM, 9-Mar-1997.)

Assertion

Ref	Expression
freq1	⊢ (𝑅 = 𝑆 → (𝑅 Fr 𝐴 ↔ 𝑆 Fr 𝐴))

Proof of Theorem freq1

Dummy variable 𝑠 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		frforeq1 4181	. . 3 ⊢ (𝑅 = 𝑆 → ( FrFor 𝑅𝐴𝑠 ↔ FrFor 𝑆𝐴𝑠))
2	1	albidv 1753	. 2 ⊢ (𝑅 = 𝑆 → (∀𝑠 FrFor 𝑅𝐴𝑠 ↔ ∀𝑠 FrFor 𝑆𝐴𝑠))
3		df-frind 4170	. 2 ⊢ (𝑅 Fr 𝐴 ↔ ∀𝑠 FrFor 𝑅𝐴𝑠)
4		df-frind 4170	. 2 ⊢ (𝑆 Fr 𝐴 ↔ ∀𝑠 FrFor 𝑆𝐴𝑠)
5	2, 3, 4	3bitr4g 222	1 ⊢ (𝑅 = 𝑆 → (𝑅 Fr 𝐴 ↔ 𝑆 Fr 𝐴))

Syntax hints:   → wi 4   ↔ wb 104  ∀wal 1288   = wceq 1290   FrFor wfrfor 4165   Fr wfr 4166

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-5 1382  ax-gen 1384  ax-ie1 1428  ax-ie2 1429  ax-4 1446  ax-17 1465  ax-ial 1473  ax-ext 2071

This theorem depends on definitions:  df-bi 116  df-nf 1396  df-cleq 2082  df-clel 2085  df-ral 2365  df-br 3854  df-frfor 4169  df-frind 4170

This theorem is referenced by:  weeq1  4194