Theorem ofreq 7620

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 5095	. . . 4 ⊢ (𝑅 = 𝑆 → ((𝑓‘𝑥)𝑅(𝑔‘𝑥) ↔ (𝑓‘𝑥)𝑆(𝑔‘𝑥)))
2	1	ralbidv 3156	. . 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 5159	. 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 7617	. 2 ⊢ ∘r 𝑅 = {⟨𝑓, 𝑔⟩ ∣ ∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓‘𝑥)𝑅(𝑔‘𝑥)}
5		df-ofr 7617	. 2 ⊢ ∘r 𝑆 = {⟨𝑓, 𝑔⟩ ∣ ∀𝑥 ∈ (dom 𝑓 ∩ dom 𝑔)(𝑓‘𝑥)𝑆(𝑔‘𝑥)}
6	3, 4, 5	3eqtr4g 2793	1 ⊢ (𝑅 = 𝑆 → ∘r 𝑅 = ∘r 𝑆)

Syntax hints:   → wi 4   = wceq 1541  ∀wral 3048   ∩ cin 3897   class class class wbr 5093  {copab 5155  dom cdm 5619  ‘cfv 6486   ∘_r cofr 7615

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2115  ax-9 2123  ax-ext 2705

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1781  df-sb 2068  df-clab 2712  df-cleq 2725  df-clel 2808  df-ral 3049  df-br 5094  df-opab 5156  df-ofr 7617

This theorem is referenced by: (None)