Metamath Proof Explorer

Theorem ss2rabd

Description: Subclass of a restricted class abstraction (deduction form). Saves ax-10 , ax-11 , ax-12 over using ss2rab and sylibr . (Contributed by SN, 4-Feb-2026)

		Ref	Expression
	Hypothesis	ss2rabd.1	$⊢ φ \to \forall x \in A (ψ \to χ)$
	Assertion	ss2rabd	$⊢ φ \to \{x \in A \| ψ\} \subseteq \{x \in A \| χ\}$

Proof

Step	Hyp	Ref	Expression
1		ss2rabd.1	$⊢ φ \to \forall x \in A (ψ \to χ)$
2		df-ral	$⊢ \forall x \in A (ψ \to χ) \leftrightarrow \forall x (x \in A \to (ψ \to χ))$
3		imdistan	$⊢ (x \in A \to (ψ \to χ)) \leftrightarrow ((x \in A \land ψ) \to (x \in A \land χ))$
4	3	albii	$⊢ \forall x (x \in A \to (ψ \to χ)) \leftrightarrow \forall x ((x \in A \land ψ) \to (x \in A \land χ))$
5	2 4	bitri	$⊢ \forall x \in A (ψ \to χ) \leftrightarrow \forall x ((x \in A \land ψ) \to (x \in A \land χ))$
6	1 5	sylib	$⊢ φ \to \forall x ((x \in A \land ψ) \to (x \in A \land χ))$
7		ss2abim	$⊢ \forall x ((x \in A \land ψ) \to (x \in A \land χ)) \to \{x \| (x \in A \land ψ)\} \subseteq \{x \| (x \in A \land χ)\}$
8	6 7	syl	$⊢ φ \to \{x \| (x \in A \land ψ)\} \subseteq \{x \| (x \in A \land χ)\}$
9		df-rab	$⊢ \{x \in A \| ψ\} = \{x \| (x \in A \land ψ)\}$
10		df-rab	$⊢ \{x \in A \| χ\} = \{x \| (x \in A \land χ)\}$
11	8 9 10	3sstr4g	$⊢ φ \to \{x \in A \| ψ\} \subseteq \{x \in A \| χ\}$