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Theorem ofreq 5715
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 3766 . . . 4 (𝑅 = 𝑆 → ((𝑓𝑥)𝑅(𝑔𝑥) ↔ (𝑓𝑥)𝑆(𝑔𝑥)))
21ralbidv 2326 . . 3 (𝑅 = 𝑆 → (∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓𝑥)𝑅(𝑔𝑥) ↔ ∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓𝑥)𝑆(𝑔𝑥)))
32opabbidv 3823 . 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 5713 . 2 𝑟 𝑅 = {⟨𝑓, 𝑔⟩ ∣ ∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓𝑥)𝑅(𝑔𝑥)}
5 df-ofr 5713 . 2 𝑟 𝑆 = {⟨𝑓, 𝑔⟩ ∣ ∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓𝑥)𝑆(𝑔𝑥)}
63, 4, 53eqtr4g 2097 1 (𝑅 = 𝑆 → ∘𝑟 𝑅 = ∘𝑟 𝑆)
Colors of variables: wff set class
Syntax hints:  wi 4   = wceq 1243  wral 2306  cin 2916   class class class wbr 3764  {copab 3817  dom cdm 4345  cfv 4902  𝑟 cofr 5711
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-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-11 1397  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022
This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-ral 2311  df-br 3765  df-opab 3819  df-ofr 5713
This theorem is referenced by: (None)
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