Theorem freq2 4101

Description: Equality theorem for the well-founded predicate. (Contributed by NM, 3-Apr-1994.)

Assertion

Ref	Expression
freq2	⊢ (𝐴 = 𝐵 → (𝑅 Fr 𝐴 ↔ 𝑅 Fr 𝐵))

Proof of Theorem freq2

Dummy variable 𝑠 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		frforeq2 4100	. . 3 ⊢ (𝐴 = 𝐵 → ( FrFor 𝑅𝐴𝑠 ↔ FrFor 𝑅𝐵𝑠))
2	1	albidv 1745	. 2 ⊢ (𝐴 = 𝐵 → (∀𝑠 FrFor 𝑅𝐴𝑠 ↔ ∀𝑠 FrFor 𝑅𝐵𝑠))
3		df-frind 4087	. 2 ⊢ (𝑅 Fr 𝐴 ↔ ∀𝑠 FrFor 𝑅𝐴𝑠)
4		df-frind 4087	. 2 ⊢ (𝑅 Fr 𝐵 ↔ ∀𝑠 FrFor 𝑅𝐵𝑠)
5	2, 3, 4	3bitr4g 221	1 ⊢ (𝐴 = 𝐵 → (𝑅 Fr 𝐴 ↔ 𝑅 Fr 𝐵))

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-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-ral 2353  df-in 2979  df-ss 2986  df-frfor 4086  df-frind 4087

This theorem is referenced by:  weeq2  4112