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Theorem ofreq 7065
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 4806 . . . 4 (𝑅 = 𝑆 → ((𝑓𝑥)𝑅(𝑔𝑥) ↔ (𝑓𝑥)𝑆(𝑔𝑥)))
21ralbidv 3124 . . 3 (𝑅 = 𝑆 → (∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓𝑥)𝑅(𝑔𝑥) ↔ ∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓𝑥)𝑆(𝑔𝑥)))
32opabbidv 4868 . 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 7063 . 2 𝑟 𝑅 = {⟨𝑓, 𝑔⟩ ∣ ∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓𝑥)𝑅(𝑔𝑥)}
5 df-ofr 7063 . 2 𝑟 𝑆 = {⟨𝑓, 𝑔⟩ ∣ ∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓𝑥)𝑆(𝑔𝑥)}
63, 4, 53eqtr4g 2819 1 (𝑅 = 𝑆 → ∘𝑟 𝑅 = ∘𝑟 𝑆)
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1632  wral 3050  cin 3714   class class class wbr 4804  {copab 4864  dom cdm 5266  cfv 6049  𝑟 cofr 7061
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1871  ax-4 1886  ax-5 1988  ax-6 2054  ax-7 2090  ax-9 2148  ax-10 2168  ax-11 2183  ax-12 2196  ax-13 2391  ax-ext 2740
This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-tru 1635  df-ex 1854  df-nf 1859  df-sb 2047  df-clab 2747  df-cleq 2753  df-clel 2756  df-ral 3055  df-br 4805  df-opab 4865  df-ofr 7063
This theorem is referenced by: (None)
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