Theorem sbc7

Description: An equivalence for class substitution in the spirit of df-clab . Note that x and A don't have to be distinct. (Contributed by NM, 18-Nov-2008) (Revised by Mario Carneiro, 13-Oct-2016)

	Ref	Expression
Assertion	sbc7	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi \leftrightarrow \exists y\left(y=A\wedge \underset{\dot{}}{[}y/x\underset{\dot{}}{]}\phi \right)$$

Proof

Step	Hyp	Ref	Expression
1		sbccow	$$⊢\underset{\dot{}}{[}A/y\underset{\dot{}}{]}\underset{\dot{}}{[}y/x\underset{\dot{}}{]}\phi \leftrightarrow \underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi$$
2		sbc5	$$⊢\underset{\dot{}}{[}A/y\underset{\dot{}}{]}\underset{\dot{}}{[}y/x\underset{\dot{}}{]}\phi \leftrightarrow \exists y\left(y=A\wedge \underset{\dot{}}{[}y/x\underset{\dot{}}{]}\phi \right)$$
3	1 2	bitr3i	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi \leftrightarrow \exists y\left(y=A\wedge \underset{\dot{}}{[}y/x\underset{\dot{}}{]}\phi \right)$$