Theorem scotteqd

Description: Equality theorem for the Scott operation. (Contributed by Rohan Ridenour, 9-Aug-2023)

	Ref	Expression
Hypothesis	scotteqd.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion	scotteqd	⊢ ( 𝜑 → Scott 𝐴 = Scott 𝐵 )

Proof

Step	Hyp	Ref	Expression
1		scotteqd.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
2	1	adantr	⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → 𝐴 = 𝐵 )
3	2	raleqdv	⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → ( ∀ 𝑦 ∈ 𝐴 ( rank ‘ 𝑥 ) ⊆ ( rank ‘ 𝑦 ) ↔ ∀ 𝑦 ∈ 𝐵 ( rank ‘ 𝑥 ) ⊆ ( rank ‘ 𝑦 ) ) )
4	1 3	rabeqbidva	⊢ ( 𝜑 → { 𝑥 ∈ 𝐴 ∣ ∀ 𝑦 ∈ 𝐴 ( rank ‘ 𝑥 ) ⊆ ( rank ‘ 𝑦 ) } = { 𝑥 ∈ 𝐵 ∣ ∀ 𝑦 ∈ 𝐵 ( rank ‘ 𝑥 ) ⊆ ( rank ‘ 𝑦 ) } )
5		df-scott	⊢ Scott 𝐴 = { 𝑥 ∈ 𝐴 ∣ ∀ 𝑦 ∈ 𝐴 ( rank ‘ 𝑥 ) ⊆ ( rank ‘ 𝑦 ) }
6		df-scott	⊢ Scott 𝐵 = { 𝑥 ∈ 𝐵 ∣ ∀ 𝑦 ∈ 𝐵 ( rank ‘ 𝑥 ) ⊆ ( rank ‘ 𝑦 ) }
7	4 5 6	3eqtr4g	⊢ ( 𝜑 → Scott 𝐴 = Scott 𝐵 )