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 3741	. 2 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥] ¬ 𝜑 ↔ [𝐴 / 𝑥] ¬ 𝜑))
2		dfsbcq2 3741	. . 3 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜑))
3	2	notbid 321	. 2 ⊢ (𝑦 = 𝐴 → (¬ [𝑦 / 𝑥]𝜑 ↔ ¬ [𝐴 / 𝑥]𝜑))
4		sbn 2313	. 2 ⊢ ([𝑦 / 𝑥] ¬ 𝜑 ↔ ¬ [𝑦 / 𝑥]𝜑)
5	1, 3, 4	vtoclbg 3519	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥] ¬ 𝜑 ↔ ¬ [𝐴 / 𝑥]𝜑))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   = wceq 1569  [wsb 2098   ∈ wcel 2144  [wsbc 3738

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-10 2177  ax-12 2212  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-nf 1816  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-sbc 3739

This theorem is used by:  sbcn1  3790  sbcrext  3820  sbcnel12g  4371  sbcne12  4372  bnj23  35397  bnj110  35536  bnj1204  35690  sbcni  39150  rspcsbnea  43288  frege124d  44868  onfrALTlem5  45632  onfrALTlem5VD  45974