Theorem freq1 4081

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 4080	. . 3 ⊢ (𝑅 = 𝑆 → ( FrFor 𝑅𝐴𝑠 ↔ FrFor 𝑆𝐴𝑠))
2	1	albidv 1705	. 2 ⊢ (𝑅 = 𝑆 → (∀𝑠 FrFor 𝑅𝐴𝑠 ↔ ∀𝑠 FrFor 𝑆𝐴𝑠))
3		df-frind 4069	. 2 ⊢ (𝑅 Fr 𝐴 ↔ ∀𝑠 FrFor 𝑅𝐴𝑠)
4		df-frind 4069	. 2 ⊢ (𝑆 Fr 𝐴 ↔ ∀𝑠 FrFor 𝑆𝐴𝑠)
5	2, 3, 4	3bitr4g 212	1 ⊢ (𝑅 = 𝑆 → (𝑅 Fr 𝐴 ↔ 𝑆 Fr 𝐴))

Syntax hints:   → wi 4   ↔ wb 98  ∀wal 1241   = wceq 1243   FrFor wfrfor 4064   Fr wfr 4065

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1336  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-4 1400  ax-17 1419  ax-ial 1427  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-nf 1350  df-cleq 2033  df-clel 2036  df-ral 2311  df-br 3765  df-frfor 4068  df-frind 4069

This theorem is referenced by:  weeq1  4093