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Theorem ofreq 7680
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 5112 . . . 4 (𝑅 = 𝑆 → ((𝑓𝑥)𝑅(𝑔𝑥) ↔ (𝑓𝑥)𝑆(𝑔𝑥)))
21ralbidv 3188 . . 3 (𝑅 = 𝑆 → (∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓𝑥)𝑅(𝑔𝑥) ↔ ∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓𝑥)𝑆(𝑔𝑥)))
32opabbidv 5178 . 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 7677 . 2 r 𝑅 = {⟨𝑓, 𝑔⟩ ∣ ∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓𝑥)𝑅(𝑔𝑥)}
5 df-ofr 7677 . 2 r 𝑆 = {⟨𝑓, 𝑔⟩ ∣ ∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓𝑥)𝑆(𝑔𝑥)}
63, 4, 53eqtr4g 2823 1 (𝑅 = 𝑆 → ∘r 𝑅 = ∘r 𝑆)
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1570  wral 3079  cin 3905   class class class wbr 5110  {copab 5174  dom cdm 5663  cfv 6538  r cofr 7675
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-ext 2735
This theorem depends on definitions:  df-bi 210  df-an 401  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-ral 3080  df-br 5111  df-opab 5175  df-ofr 7677
This theorem is referenced by: (None)
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