Theorem scotteqd

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

	Ref	Expression
Hypothesis	scotteqd.1	|- ( ph -> A = B )
Assertion	scotteqd	|- ( ph -> Scott A = Scott B )

Proof

Step	Hyp	Ref	Expression
1		scotteqd.1	|- ( ph -> A = B )
2	1	adantr	|- ( ( ph /\ x e. A ) -> A = B )
3	2	raleqdv	|- ( ( ph /\ x e. A ) -> ( A. y e. A ( rank ` x ) C_ ( rank ` y ) <-> A. y e. B ( rank ` x ) C_ ( rank ` y ) ) )
4	1 3	rabeqbidva	|- ( ph -> { x e. A | A. y e. A ( rank ` x ) C_ ( rank ` y ) } = { x e. B | A. y e. B ( rank ` x ) C_ ( rank ` y ) } )
5		df-scott	|- Scott A = { x e. A | A. y e. A ( rank ` x ) C_ ( rank ` y ) }
6		df-scott	|- Scott B = { x e. B | A. y e. B ( rank ` x ) C_ ( rank ` y ) }
7	4 5 6	3eqtr4g	|- ( ph -> Scott A = Scott B )