Theorem sbcn1

Description: Move negation in and out of class substitution. One direction of sbcng that holds for proper classes. (Contributed by NM, 17-Aug-2018)

	Ref	Expression
Assertion	sbcn1	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\neg \phi \to \neg \underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi$$

Proof

Step	Hyp	Ref	Expression
1		sbcex	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\neg \phi \to A\in \mathrm{V}$$
2		sbcng	$$⊢A\in \mathrm{V}\to \left(\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\neg \phi \leftrightarrow \neg \underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi \right)$$
3	2	biimpd	$$⊢A\in \mathrm{V}\to \left(\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\neg \phi \to \neg \underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi \right)$$
4	1 3	mpcom	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\neg \phi \to \neg \underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi$$