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Theorem sbc6g 2839
Description: An equivalence for class substitution. (Contributed by NM, 11-Oct-2004.) (Proof shortened by Andrew Salmon, 8-Jun-2011.)
Assertion
Ref Expression
sbc6g (𝐴𝑉 → ([𝐴 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝐴𝜑)))
Distinct variable group:   𝑥,𝐴
Allowed substitution hints:   𝜑(𝑥)   𝑉(𝑥)

Proof of Theorem sbc6g
StepHypRef Expression
1 nfe1 1425 . . 3 𝑥𝑥(𝑥 = 𝐴𝜑)
2 ceqex 2722 . . 3 (𝑥 = 𝐴 → (𝜑 ↔ ∃𝑥(𝑥 = 𝐴𝜑)))
31, 2ceqsalg 2627 . 2 (𝐴𝑉 → (∀𝑥(𝑥 = 𝐴𝜑) ↔ ∃𝑥(𝑥 = 𝐴𝜑)))
4 sbc5 2838 . 2 ([𝐴 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝐴𝜑))
53, 4syl6rbbr 197 1 (𝐴𝑉 → ([𝐴 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝐴𝜑)))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 102  wb 103  wal 1282   = wceq 1284  wex 1421  wcel 1433  [wsbc 2815
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063
This theorem depends on definitions:  df-bi 115  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-v 2603  df-sbc 2816
This theorem is referenced by:  sbc6  2840  sbciegft  2844  ralsnsg  3430  ralsns  3431  fz1sbc  9113
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