Theorem sbcng 3835

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 3790	. 2 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥] ¬ 𝜑 ↔ [𝐴 / 𝑥] ¬ 𝜑))
2		dfsbcq2 3790	. . 3 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜑))
3	2	notbid 318	. 2 ⊢ (𝑦 = 𝐴 → (¬ [𝑦 / 𝑥]𝜑 ↔ ¬ [𝐴 / 𝑥]𝜑))
4		sbn 2279	. 2 ⊢ ([𝑦 / 𝑥] ¬ 𝜑 ↔ ¬ [𝑦 / 𝑥]𝜑)
5	1, 3, 4	vtoclbg 3556	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥] ¬ 𝜑 ↔ ¬ [𝐴 / 𝑥]𝜑))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 206   = wceq 1539  [wsb 2063   ∈ wcel 2107  [wsbc 3787

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808  ax-5 1909  ax-6 1966  ax-7 2006  ax-8 2109  ax-9 2117  ax-10 2140  ax-12 2176  ax-ext 2707

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1542  df-ex 1779  df-nf 1783  df-sb 2064  df-clab 2714  df-cleq 2728  df-clel 2815  df-sbc 3788

This theorem is referenced by:  sbcn1  3840  sbcrext  3872  sbcnel12g  4413  sbcne12  4414  difopabOLD  5840  bnj23  34733  bnj110  34873  bnj1204  35027  sbcni  38119  rspcsbnea  42133  frege124d  43779  onfrALTlem5  44567  onfrALTlem5VD  44910