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Theorem ofreq 6853
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 4615 . . . 4 (𝑅 = 𝑆 → ((𝑓𝑥)𝑅(𝑔𝑥) ↔ (𝑓𝑥)𝑆(𝑔𝑥)))
21ralbidv 2980 . . 3 (𝑅 = 𝑆 → (∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓𝑥)𝑅(𝑔𝑥) ↔ ∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓𝑥)𝑆(𝑔𝑥)))
32opabbidv 4678 . 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 6851 . 2 𝑟 𝑅 = {⟨𝑓, 𝑔⟩ ∣ ∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓𝑥)𝑅(𝑔𝑥)}
5 df-ofr 6851 . 2 𝑟 𝑆 = {⟨𝑓, 𝑔⟩ ∣ ∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓𝑥)𝑆(𝑔𝑥)}
63, 4, 53eqtr4g 2680 1 (𝑅 = 𝑆 → ∘𝑟 𝑅 = ∘𝑟 𝑆)
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1480  wral 2907  cin 3554   class class class wbr 4613  {copab 4672  dom cdm 5074  cfv 5847  𝑟 cofr 6849
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734  ax-5 1836  ax-6 1885  ax-7 1932  ax-9 1996  ax-10 2016  ax-11 2031  ax-12 2044  ax-13 2245  ax-ext 2601
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1483  df-ex 1702  df-nf 1707  df-sb 1878  df-clab 2608  df-cleq 2614  df-clel 2617  df-ral 2912  df-br 4614  df-opab 4674  df-ofr 6851
This theorem is referenced by: (None)
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