Theorem ofreq 7698

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 5104	. . . 4 ⊢ (𝑅 = 𝑆 → ((𝑓‘𝑥)𝑅(𝑔‘𝑥) ↔ (𝑓‘𝑥)𝑆(𝑔‘𝑥)))
2	1	ralbidv 3185	. . 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 5170	. 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 7695	. 2 ⊢ ∘r 𝑅 = {⟨𝑓, 𝑔⟩ ∣ ∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓‘𝑥)𝑅(𝑔‘𝑥)}
5		df-ofr 7695	. 2 ⊢ ∘r 𝑆 = {⟨𝑓, 𝑔⟩ ∣ ∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓‘𝑥)𝑆(𝑔‘𝑥)}
6	3, 4, 5	3eqtr4g 2820	1 ⊢ (𝑅 = 𝑆 → ∘r 𝑅 = ∘r 𝑆)

This proof depends on syntax axioms:   → wi 4   = wceq 1569  ∀wral 3076   ∩ cin 4089   class class class wbr 5102  {copab 5166  dom cdm 5650  ‘cfv 6539   ∘_r cofr 7693

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-br 5103  df-opab 5167  df-ofr 7695

This theorem is used by: (None)