Metamath Proof Explorer


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 ( [ 𝐴 / 𝑥 ] ¬ 𝜑 → ¬ [ 𝐴 / 𝑥 ] 𝜑 )

Proof

Step Hyp Ref Expression
1 sbcex ( [ 𝐴 / 𝑥 ] ¬ 𝜑𝐴 ∈ V )
2 sbcng ( 𝐴 ∈ V → ( [ 𝐴 / 𝑥 ] ¬ 𝜑 ↔ ¬ [ 𝐴 / 𝑥 ] 𝜑 ) )
3 2 biimpd ( 𝐴 ∈ V → ( [ 𝐴 / 𝑥 ] ¬ 𝜑 → ¬ [ 𝐴 / 𝑥 ] 𝜑 ) )
4 1 3 mpcom ( [ 𝐴 / 𝑥 ] ¬ 𝜑 → ¬ [ 𝐴 / 𝑥 ] 𝜑 )