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Theorem sbc6g 2928
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 1472 . . 3 𝑥𝑥(𝑥 = 𝐴𝜑)
2 ceqex 2807 . . 3 (𝑥 = 𝐴 → (𝜑 ↔ ∃𝑥(𝑥 = 𝐴𝜑)))
31, 2ceqsalg 2709 . 2 (𝐴𝑉 → (∀𝑥(𝑥 = 𝐴𝜑) ↔ ∃𝑥(𝑥 = 𝐴𝜑)))
4 sbc5 2927 . 2 ([𝐴 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝐴𝜑))
53, 4syl6rbbr 198 1 (𝐴𝑉 → ([𝐴 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝐴𝜑)))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 103  wb 104  wal 1329   = wceq 1331  wex 1468  wcel 1480  [wsbc 2904
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2119
This theorem depends on definitions:  df-bi 116  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2124  df-cleq 2130  df-clel 2133  df-nfc 2268  df-v 2683  df-sbc 2905
This theorem is referenced by:  sbc6  2929  sbciegft  2934  ralsnsg  3556  ralsns  3557  fz1sbc  9869
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