Theorem eqabcbw

Description: Version of eqabcb using implicit substitution, which requires fewer axioms. (Contributed by TM, 24-Jan-2026)

	Ref	Expression
Hypothesis	eqabbw.1	$$⊢x=y\to \left(\phi \leftrightarrow \psi \right)$$
Assertion	eqabcbw	$$⊢\left\{x|\phi \right\}=A\leftrightarrow \forall y\left(\psi \leftrightarrow y\in A\right)$$

Proof

Step	Hyp	Ref	Expression
1		eqabbw.1	$$⊢x=y\to \left(\phi \leftrightarrow \psi \right)$$
2	1	eqabbw	$$⊢A=\left\{x|\phi \right\}\leftrightarrow \forall y\left(y\in A\leftrightarrow \psi \right)$$
3		eqcom	$$⊢\left\{x|\phi \right\}=A\leftrightarrow A=\left\{x|\phi \right\}$$
4		bicom	$$⊢\left(\psi \leftrightarrow y\in A\right)\leftrightarrow \left(y\in A\leftrightarrow \psi \right)$$
5	4	albii	$$⊢\forall y\left(\psi \leftrightarrow y\in A\right)\leftrightarrow \forall y\left(y\in A\leftrightarrow \psi \right)$$
6	2 3 5	3bitr4i	$$⊢\left\{x|\phi \right\}=A\leftrightarrow \forall y\left(\psi \leftrightarrow y\in A\right)$$