Theorem sbc5

Description: An equivalence for class substitution. (Contributed by NM, 23-Aug-1993) (Revised by Mario Carneiro, 12-Oct-2016) (Proof shortened by SN, 2-Sep-2024)

	Ref	Expression
Assertion	sbc5	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi \leftrightarrow \exists x\left(x=A\wedge \phi \right)$$

Proof

Step	Hyp	Ref	Expression
1		df-sbc	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi \leftrightarrow A\in \left\{x|\phi \right\}$$
2		clelab	$$⊢A\in \left\{x|\phi \right\}\leftrightarrow \exists x\left(x=A\wedge \phi \right)$$
3	1 2	bitri	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi \leftrightarrow \exists x\left(x=A\wedge \phi \right)$$