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Theorem ofreq 5657
Description: Equality theorem for function relation. (Contributed by Mario Carneiro, 28-Jul-2014.)
Assertion
Ref Expression
ofreq (𝑅 = 𝑆 → ∘𝑟 𝑅 = ∘𝑟 𝑆)

Proof of Theorem ofreq
Dummy variables f g x are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 breq 3757 . . . 4 (𝑅 = 𝑆 → ((fx)𝑅(gx) ↔ (fx)𝑆(gx)))
21ralbidv 2320 . . 3 (𝑅 = 𝑆 → (x (dom f ∩ dom g)(fx)𝑅(gx) ↔ x (dom f ∩ dom g)(fx)𝑆(gx)))
32opabbidv 3814 . 2 (𝑅 = 𝑆 → {⟨f, g⟩ ∣ x (dom f ∩ dom g)(fx)𝑅(gx)} = {⟨f, g⟩ ∣ x (dom f ∩ dom g)(fx)𝑆(gx)})
4 df-ofr 5655 . 2 𝑟 𝑅 = {⟨f, g⟩ ∣ x (dom f ∩ dom g)(fx)𝑅(gx)}
5 df-ofr 5655 . 2 𝑟 𝑆 = {⟨f, g⟩ ∣ x (dom f ∩ dom g)(fx)𝑆(gx)}
63, 4, 53eqtr4g 2094 1 (𝑅 = 𝑆 → ∘𝑟 𝑅 = ∘𝑟 𝑆)
Colors of variables: wff set class
Syntax hints:  wi 4   = wceq 1242  wral 2300  cin 2910   class class class wbr 3755  {copab 3808  dom cdm 4288  cfv 4845  𝑟 cofr 5653
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 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-11 1394  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019
This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-ral 2305  df-br 3756  df-opab 3810  df-ofr 5655
This theorem is referenced by: (None)
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