Theorem sbcng 3637

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 3599	. 2 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥] ¬ 𝜑 ↔ [𝐴 / 𝑥] ¬ 𝜑))
2		dfsbcq2 3599	. . 3 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜑))
3	2	notbid 309	. 2 ⊢ (𝑦 = 𝐴 → (¬ [𝑦 / 𝑥]𝜑 ↔ ¬ [𝐴 / 𝑥]𝜑))
4		sbn 2482	. 2 ⊢ ([𝑦 / 𝑥] ¬ 𝜑 ↔ ¬ [𝑦 / 𝑥]𝜑)
5	1, 3, 4	vtoclbg 3419	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥] ¬ 𝜑 ↔ ¬ [𝐴 / 𝑥]𝜑))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 197   = wceq 1652  [wsb 2062   ∈ wcel 2155  [wsbc 3596

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1890  ax-4 1904  ax-5 2005  ax-6 2070  ax-7 2105  ax-9 2164  ax-10 2183  ax-12 2211  ax-13 2352  ax-ext 2743

This theorem depends on definitions:  df-bi 198  df-an 385  df-or 874  df-tru 1656  df-ex 1875  df-nf 1879  df-sb 2063  df-clab 2752  df-cleq 2758  df-clel 2761  df-v 3352  df-sbc 3597

This theorem is referenced by:  sbcn1  3642  sbcrext  3670  sbcnel12g  4146  sbcne12  4147  difopab  5422  bnj23  31235  bnj110  31376  bnj1204  31528  sbcni  34336  frege124d  38728  onfrALTlem5  39420  onfrALTlem5VD  39773