Theorem ofreq

Description: Equality theorem for function relation. (Contributed by Mario Carneiro, 28-Jul-2014)

	Ref	Expression
Assertion	ofreq	⊢ ( 𝑅 = 𝑆 → ∘r 𝑅 = ∘r 𝑆 )

Proof

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