Theorem sbcng 3818

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

Assertion

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

Proof of Theorem sbcng

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		dfsbcq2 3773	. 2 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥] ¬ 𝜑 ↔ [𝐴 / 𝑥] ¬ 𝜑))
2		dfsbcq2 3773	. . 3 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜑))
3	2	notbid 318	. 2 ⊢ (𝑦 = 𝐴 → (¬ [𝑦 / 𝑥]𝜑 ↔ ¬ [𝐴 / 𝑥]𝜑))
4		sbn 2281	. 2 ⊢ ([𝑦 / 𝑥] ¬ 𝜑 ↔ ¬ [𝑦 / 𝑥]𝜑)
5	1, 3, 4	vtoclbg 3541	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥] ¬ 𝜑 ↔ ¬ [𝐴 / 𝑥]𝜑))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 206   = wceq 1540  [wsb 2065   ∈ wcel 2109  [wsbc 3770

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-10 2142  ax-12 2178  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1543  df-ex 1780  df-nf 1784  df-sb 2066  df-clab 2715  df-cleq 2728  df-clel 2810  df-sbc 3771

This theorem is referenced by:  sbcn1  3823  sbcrext  3853  sbcnel12g  4394  sbcne12  4395  difopabOLD  5815  bnj23  34754  bnj110  34894  bnj1204  35048  sbcni  38140  rspcsbnea  42149  frege124d  43752  onfrALTlem5  44534  onfrALTlem5VD  44876