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Metamath Proof Explorer

Theorem eqab

Description: One direction of eqabb is provable from fewer axioms. (Contributed by Wolf Lammen, 13-Feb-2025)

		Ref	Expression
	Assertion	eqab	$⊢ \forall x (x \in A \leftrightarrow φ) \to A = \{x \| φ\}$

Step	Hyp	Ref	Expression
1		abid1	$⊢ A = \{x \| x \in A\}$
2		abbi	$⊢ \forall x (x \in A \leftrightarrow φ) \to \{x \| x \in A\} = \{x \| φ\}$
3	1 2	eqtrid	$⊢ \forall x (x \in A \leftrightarrow φ) \to A = \{x \| φ\}$