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Theorem ofreq 6085
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 4005 . . . 4 (𝑅 = 𝑆 → ((𝑓𝑥)𝑅(𝑔𝑥) ↔ (𝑓𝑥)𝑆(𝑔𝑥)))
21ralbidv 2477 . . 3 (𝑅 = 𝑆 → (∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓𝑥)𝑅(𝑔𝑥) ↔ ∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓𝑥)𝑆(𝑔𝑥)))
32opabbidv 4069 . 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 6083 . 2 𝑟 𝑅 = {⟨𝑓, 𝑔⟩ ∣ ∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓𝑥)𝑅(𝑔𝑥)}
5 df-ofr 6083 . 2 𝑟 𝑆 = {⟨𝑓, 𝑔⟩ ∣ ∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓𝑥)𝑆(𝑔𝑥)}
63, 4, 53eqtr4g 2235 1 (𝑅 = 𝑆 → ∘𝑟 𝑅 = ∘𝑟 𝑆)
Colors of variables: wff set class
Syntax hints:  wi 4   = wceq 1353  wral 2455  cin 3128   class class class wbr 4003  {copab 4063  dom cdm 4626  cfv 5216  𝑟 cofr 6081
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 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-11 1506  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159
This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-ral 2460  df-br 4004  df-opab 4065  df-ofr 6083
This theorem is referenced by: (None)
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