Theorem sbcng 3785

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 3740	. 2 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥] ¬ 𝜑 ↔ [𝐴 / 𝑥] ¬ 𝜑))
2		dfsbcq2 3740	. . 3 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜑))
3	2	notbid 318	. 2 ⊢ (𝑦 = 𝐴 → (¬ [𝑦 / 𝑥]𝜑 ↔ ¬ [𝐴 / 𝑥]𝜑))
4		sbn 2284	. 2 ⊢ ([𝑦 / 𝑥] ¬ 𝜑 ↔ ¬ [𝑦 / 𝑥]𝜑)
5	1, 3, 4	vtoclbg 3511	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥] ¬ 𝜑 ↔ ¬ [𝐴 / 𝑥]𝜑))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 206   = wceq 1541  [wsb 2067   ∈ wcel 2113  [wsbc 3737

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2115  ax-9 2123  ax-10 2146  ax-12 2182  ax-ext 2705

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1544  df-ex 1781  df-nf 1785  df-sb 2068  df-clab 2712  df-cleq 2725  df-clel 2808  df-sbc 3738

This theorem is referenced by:  sbcn1  3790  sbcrext  3820  sbcnel12g  4363  sbcne12  4364  bnj23  34753  bnj110  34893  bnj1204  35047  sbcni  38174  rspcsbnea  42247  frege124d  43881  onfrALTlem5  44662  onfrALTlem5VD  45004