Theorem rabss2

Description: Subclass law for restricted abstraction. (Contributed by NM, 18-Dec-2004) (Proof shortened by Andrew Salmon, 26-Jun-2011) Avoid axioms. (Revised by TM, 1-Feb-2026)

	Ref	Expression
Assertion	rabss2	$$⊢A\subseteq B\to \left\{x\in A|\phi \right\}\subseteq \left\{x\in B|\phi \right\}$$

Proof

Step	Hyp	Ref	Expression
1		ssel	$$⊢A\subseteq B\to \left(x\in A\to x\in B\right)$$
2	1	anim1d	$$⊢A\subseteq B\to \left(\left(x\in A\wedge \phi \right)\to \left(x\in B\wedge \phi \right)\right)$$
3	2	ss2abdv	$$⊢A\subseteq B\to \left\{x|\left(x\in A\wedge \phi \right)\right\}\subseteq \left\{x|\left(x\in B\wedge \phi \right)\right\}$$
4		df-rab	$$⊢\left\{x\in A|\phi \right\}=\left\{x|\left(x\in A\wedge \phi \right)\right\}$$
5		df-rab	$$⊢\left\{x\in B|\phi \right\}=\left\{x|\left(x\in B\wedge \phi \right)\right\}$$
6	3 4 5	3sstr4g	$$⊢A\subseteq B\to \left\{x\in A|\phi \right\}\subseteq \left\{x\in B|\phi \right\}$$