Theorem freq1 4099

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 4098	. . 3 ⊢ (𝑅 = 𝑆 → ( FrFor 𝑅𝐴𝑠 ↔ FrFor 𝑆𝐴𝑠))
2	1	albidv 1745	. 2 ⊢ (𝑅 = 𝑆 → (∀𝑠 FrFor 𝑅𝐴𝑠 ↔ ∀𝑠 FrFor 𝑆𝐴𝑠))
3		df-frind 4087	. 2 ⊢ (𝑅 Fr 𝐴 ↔ ∀𝑠 FrFor 𝑅𝐴𝑠)
4		df-frind 4087	. 2 ⊢ (𝑆 Fr 𝐴 ↔ ∀𝑠 FrFor 𝑆𝐴𝑠)
5	2, 3, 4	3bitr4g 221	1 ⊢ (𝑅 = 𝑆 → (𝑅 Fr 𝐴 ↔ 𝑆 Fr 𝐴))

Syntax hints:   → wi 4   ↔ wb 103  ∀wal 1282   = wceq 1284   FrFor wfrfor 4082   Fr wfr 4083

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1376  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-4 1440  ax-17 1459  ax-ial 1467  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-nf 1390  df-cleq 2074  df-clel 2077  df-ral 2353  df-br 3786  df-frfor 4086  df-frind 4087

This theorem is referenced by:  weeq1  4111