Theorem sbcng 1972

Description: Move negation in and out of class substitution.

Assertion

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

Proof of Theorem sbcng

Step	Hyp	Ref	Expression
1		dfsbcq 1946	. 2 ⊢ (y = A → ([y / x] ¬ φ ↔ [A / x] ¬ φ))
2		dfsbcq 1946	. . 3 ⊢ (y = A → ([y / x]φ ↔ [A / x]φ))
3	2	negbid 613	. 2 ⊢ (y = A → (¬ [y / x]φ ↔ ¬ [A / x]φ))
4		sbn 1233	. 2 ⊢ ([y / x] ¬ φ ↔ ¬ [y / x]φ)
5	1, 3, 4	vtoclbg 1851	1 ⊢ (A ∈ B → ([A / x] ¬ φ ↔ ¬ [A / x]φ))

Syntax hints:  ¬ wn 2   → wi 3   ↔ wb 146   = wceq 958   ∈ wcel 960  [wsbc 1172

This theorem is referenced by:  sbcrext 1994  sbcrexgf 1996  ra4esbca 2002  rexpr 2433

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 964  ax-gen 965  ax-8 966  ax-10 968  ax-12 970  ax-17 973  ax-4 975  ax-5o 977  ax-6o 980  ax-9o 1125  ax-10o 1142  ax-11o 1220  ax-ext 1462

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 983  df-sb 1174  df-clab 1467  df-cleq 1472  df-clel 1475  df-v 1815  df-sbc 1945

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