Theorem ofreq 7691

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 5116	. . . 4 ⊢ (𝑅 = 𝑆 → ((𝑓‘𝑥)𝑅(𝑔‘𝑥) ↔ (𝑓‘𝑥)𝑆(𝑔‘𝑥)))
2	1	ralbidv 3191	. . 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 5182	. 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 7688	. 2 ⊢ ∘r 𝑅 = {⟨𝑓, 𝑔⟩ ∣ ∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓‘𝑥)𝑅(𝑔‘𝑥)}
5		df-ofr 7688	. 2 ⊢ ∘r 𝑆 = {⟨𝑓, 𝑔⟩ ∣ ∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓‘𝑥)𝑆(𝑔‘𝑥)}
6	3, 4, 5	3eqtr4g 2826	1 ⊢ (𝑅 = 𝑆 → ∘r 𝑅 = ∘r 𝑆)

This proof depends on syntax axioms:   → wi 4   = wceq 1570  ∀wral 3082   ∩ cin 3907   class class class wbr 5114  {copab 5178  dom cdm 5666  ‘cfv 6543   ∘_r cofr 7686

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2738

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-sb 2100  df-clab 2745  df-cleq 2758  df-clel 2841  df-ral 3083  df-br 5115  df-opab 5179  df-ofr 7688

This theorem is used by: (None)