Theorem scotteqd

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

	Ref	Expression
Hypothesis	scotteqd.1	$$⊢\phi \to A=B$$
Assertion	scotteqd	$$⊢\phi \to ScottA=ScottB$$

Proof

Step	Hyp	Ref	Expression
1		scotteqd.1	$$⊢\phi \to A=B$$
2	1	adantr	$$⊢\left(\phi \wedge x\in A\right)\to A=B$$
3	2	raleqdv	$$⊢\left(\phi \wedge x\in A\right)\to \left(\forall y\in A\mathrm{rank}\left(x\right)\subseteq \mathrm{rank}\left(y\right)\leftrightarrow \forall y\in B\mathrm{rank}\left(x\right)\subseteq \mathrm{rank}\left(y\right)\right)$$
4	1 3	rabeqbidva	$$⊢\phi \to \left\{x\in A|\forall y\in A\mathrm{rank}\left(x\right)\subseteq \mathrm{rank}\left(y\right)\right\}=\left\{x\in B|\forall y\in B\mathrm{rank}\left(x\right)\subseteq \mathrm{rank}\left(y\right)\right\}$$
5		df-scott	$$⊢ScottA=\left\{x\in A|\forall y\in A\mathrm{rank}\left(x\right)\subseteq \mathrm{rank}\left(y\right)\right\}$$
6		df-scott	$$⊢ScottB=\left\{x\in B|\forall y\in B\mathrm{rank}\left(x\right)\subseteq \mathrm{rank}\left(y\right)\right\}$$
7	4 5 6	3eqtr4g	$$⊢\phi \to ScottA=ScottB$$