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Theorem freq1 4398
Description: Equality theorem for the well-founded predicate. (Contributed by NM, 9-Mar-1997.)
Assertion
Ref Expression
freq1 (𝑅 = 𝑆 → (𝑅 Fr 𝐴𝑆 Fr 𝐴))

Proof of Theorem freq1
Dummy variable 𝑠 is distinct from all other variables.
StepHypRef Expression
1 frforeq1 4397 . . 3 (𝑅 = 𝑆 → ( FrFor 𝑅𝐴𝑠 ↔ FrFor 𝑆𝐴𝑠))
21albidv 1848 . 2 (𝑅 = 𝑆 → (∀𝑠 FrFor 𝑅𝐴𝑠 ↔ ∀𝑠 FrFor 𝑆𝐴𝑠))
3 df-frind 4386 . 2 (𝑅 Fr 𝐴 ↔ ∀𝑠 FrFor 𝑅𝐴𝑠)
4 df-frind 4386 . 2 (𝑆 Fr 𝐴 ↔ ∀𝑠 FrFor 𝑆𝐴𝑠)
52, 3, 43bitr4g 223 1 (𝑅 = 𝑆 → (𝑅 Fr 𝐴𝑆 Fr 𝐴))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 105  wal 1371   = wceq 1373   FrFor wfrfor 4381   Fr wfr 4382
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1471  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-4 1534  ax-17 1550  ax-ial 1558  ax-ext 2188
This theorem depends on definitions:  df-bi 117  df-nf 1485  df-cleq 2199  df-clel 2202  df-ral 2490  df-br 4051  df-frfor 4385  df-frind 4386
This theorem is referenced by:  weeq1  4410
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