Theorem sbcn1 2900

Description: Move negation in and out of class substitution. One direction of sbcng 2893 that holds for proper classes. (Contributed by NM, 17-Aug-2018.)

Assertion

Ref	Expression
sbcn1	⊢ ([𝐴 / 𝑥] ¬ 𝜑 → ¬ [𝐴 / 𝑥]𝜑)

Proof of Theorem sbcn1

Step	Hyp	Ref	Expression
1		sbcex 2862	. 2 ⊢ ([𝐴 / 𝑥] ¬ 𝜑 → 𝐴 ∈ V)
2		sbcng 2893	. . 3 ⊢ (𝐴 ∈ V → ([𝐴 / 𝑥] ¬ 𝜑 ↔ ¬ [𝐴 / 𝑥]𝜑))
3	2	biimpd 143	. 2 ⊢ (𝐴 ∈ V → ([𝐴 / 𝑥] ¬ 𝜑 → ¬ [𝐴 / 𝑥]𝜑))
4	1, 3	mpcom 36	1 ⊢ ([𝐴 / 𝑥] ¬ 𝜑 → ¬ [𝐴 / 𝑥]𝜑)

Syntax hints:  ¬ wn 3   → wi 4   ∈ wcel 1445  Vcvv 2633  [wsbc 2854

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 582  ax-in2 583  ax-io 668  ax-5 1388  ax-7 1389  ax-gen 1390  ax-ie1 1434  ax-ie2 1435  ax-8 1447  ax-10 1448  ax-11 1449  ax-i12 1450  ax-bndl 1451  ax-4 1452  ax-17 1471  ax-i9 1475  ax-ial 1479  ax-i5r 1480  ax-ext 2077

This theorem depends on definitions:  df-bi 116  df-tru 1299  df-fal 1302  df-nf 1402  df-sb 1700  df-clab 2082  df-cleq 2088  df-clel 2091  df-nfc 2224  df-v 2635  df-sbc 2855

This theorem is referenced by: (None)