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Metamath Proof Explorer

Theorem rabbida4

Description: Version of rabbidva2 with disjoint variable condition replaced by nonfreeness hypothesis. (Contributed by BJ, 27-Apr-2019)

	Ref	Expression
Hypotheses	rabbida4.nf	$⊢ Ⅎ x φ$
	rabbida4.1	$⊢ φ \to ((x \in A \land ψ) \leftrightarrow (x \in B \land χ))$
Assertion	rabbida4	$⊢ φ \to \{x \in A \| ψ\} = \{x \in B \| χ\}$

Step	Hyp	Ref	Expression
1		rabbida4.nf	$⊢ Ⅎ x φ$
2		rabbida4.1	$⊢ φ \to ((x \in A \land ψ) \leftrightarrow (x \in B \land χ))$
3	1 2	abbid	$⊢ φ \to \{x \| (x \in A \land ψ)\} = \{x \| (x \in B \land χ)\}$
4		df-rab	$⊢ \{x \in A \| ψ\} = \{x \| (x \in A \land ψ)\}$
5		df-rab	$⊢ \{x \in B \| χ\} = \{x \| (x \in B \land χ)\}$
6	3 4 5	3eqtr4g	$⊢ φ \to \{x \in A \| ψ\} = \{x \in B \| χ\}$