This is an inofficial mirror of http://metamath.tirix.org for personal testing of a visualizer extension only.

Metamath Proof Explorer

Theorem rabeqbidva

Description: Equality of restricted class abstractions. (Contributed by Mario Carneiro, 26-Jan-2017) Remove DV conditions. (Revised by GG, 1-Sep-2025)

	Ref	Expression
Hypotheses	rabeqbidva.1	$⊢ φ \to A = B$
	rabeqbidva.2	$⊢ (φ \land x \in A) \to (ψ \leftrightarrow χ)$
Assertion	rabeqbidva	$⊢ φ \to \{x \in A \| ψ\} = \{x \in B \| χ\}$

Proof

Step	Hyp	Ref	Expression
1		rabeqbidva.1	$⊢ φ \to A = B$
2		rabeqbidva.2	$⊢ (φ \land x \in A) \to (ψ \leftrightarrow χ)$
3	2	rabbidva	$⊢ φ \to \{x \in A \| ψ\} = \{x \in A \| χ\}$
4	1	eleq2d	$⊢ φ \to (x \in A \leftrightarrow x \in B)$
5	4	anbi1d	$⊢ φ \to ((x \in A \land χ) \leftrightarrow (x \in B \land χ))$
6	5	rabbidva2	$⊢ φ \to \{x \in A \| χ\} = \{x \in B \| χ\}$
7	3 6	eqtrd	$⊢ φ \to \{x \in A \| ψ\} = \{x \in B \| χ\}$