Theorem sbcnestg

Description: Nest the composition of two substitutions. Usage of this theorem is discouraged because it depends on ax-13 . Use the weaker sbcnestgw when possible. (Contributed by NM, 27-Nov-2005) (Proof shortened by Mario Carneiro, 11-Nov-2016) (New usage is discouraged.)

	Ref	Expression
Assertion	sbcnestg	$$⊢A\in V\to \left(\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\underset{\dot{}}{[}B/y\underset{\dot{}}{]}\phi \leftrightarrow \underset{\dot{}}{[}⦋A/x⦌B/y\underset{\dot{}}{]}\phi \right)$$

Proof

Step	Hyp	Ref	Expression
1		nfv	$$⊢Ⅎx\phi$$
2	1	ax-gen	$$⊢\forall yℲx\phi$$
3		sbcnestgf	$$⊢\left(A\in V\wedge \forall yℲx\phi \right)\to \left(\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\underset{\dot{}}{[}B/y\underset{\dot{}}{]}\phi \leftrightarrow \underset{\dot{}}{[}⦋A/x⦌B/y\underset{\dot{}}{]}\phi \right)$$
4	2 3	mpan2	$$⊢A\in V\to \left(\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\underset{\dot{}}{[}B/y\underset{\dot{}}{]}\phi \leftrightarrow \underset{\dot{}}{[}⦋A/x⦌B/y\underset{\dot{}}{]}\phi \right)$$