Theorem ofreq 6900

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.

Step	Hyp	Ref	Expression
1		breq 4655	. . . 4 ⊢ (𝑅 = 𝑆 → ((𝑓‘𝑥)𝑅(𝑔‘𝑥) ↔ (𝑓‘𝑥)𝑆(𝑔‘𝑥)))
2	1	ralbidv 2986	. . 3 ⊢ (𝑅 = 𝑆 → (∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓‘𝑥)𝑅(𝑔‘𝑥) ↔ ∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓‘𝑥)𝑆(𝑔‘𝑥)))
3	2	opabbidv 4716	. 2 ⊢ (𝑅 = 𝑆 → {⟨𝑓, 𝑔⟩ ∣ ∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓‘𝑥)𝑅(𝑔‘𝑥)} = {⟨𝑓, 𝑔⟩ ∣ ∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓‘𝑥)𝑆(𝑔‘𝑥)})
4		df-ofr 6898	. 2 ⊢ ∘𝑟 𝑅 = {⟨𝑓, 𝑔⟩ ∣ ∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓‘𝑥)𝑅(𝑔‘𝑥)}
5		df-ofr 6898	. 2 ⊢ ∘𝑟 𝑆 = {⟨𝑓, 𝑔⟩ ∣ ∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓‘𝑥)𝑆(𝑔‘𝑥)}
6	3, 4, 5	3eqtr4g 2681	1 ⊢ (𝑅 = 𝑆 → ∘𝑟 𝑅 = ∘𝑟 𝑆)

Syntax hints:   → wi 4   = wceq 1483  ∀wral 2912   ∩ cin 3573   class class class wbr 4653  {copab 4712  dom cdm 5114  ‘cfv 5888   ∘_𝑟 cofr 6896

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-ral 2917  df-br 4654  df-opab 4713  df-ofr 6898

This theorem is referenced by: (None)