Theorem sbcng

Description: Move negation in and out of class substitution. (Contributed by NM, 16-Jan-2004)

	Ref	Expression
Assertion	sbcng	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] ¬ 𝜑 ↔ ¬ [ 𝐴 / 𝑥 ] 𝜑 ) )

Proof

Step	Hyp	Ref	Expression
1		dfsbcq2	⊢ ( 𝑦 = 𝐴 → ( [ 𝑦 / 𝑥 ] ¬ 𝜑 ↔ [ 𝐴 / 𝑥 ] ¬ 𝜑 ) )
2		dfsbcq2	⊢ ( 𝑦 = 𝐴 → ( [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝐴 / 𝑥 ] 𝜑 ) )
3	2	notbid	⊢ ( 𝑦 = 𝐴 → ( ¬ [ 𝑦 / 𝑥 ] 𝜑 ↔ ¬ [ 𝐴 / 𝑥 ] 𝜑 ) )
4		sbn	⊢ ( [ 𝑦 / 𝑥 ] ¬ 𝜑 ↔ ¬ [ 𝑦 / 𝑥 ] 𝜑 )
5	1 3 4	vtoclbg	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] ¬ 𝜑 ↔ ¬ [ 𝐴 / 𝑥 ] 𝜑 ) )