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Theorem sbc6g 3002
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 sbc5 3001 . 2 ([𝐴 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝐴𝜑))
2 nfe1 1507 . . 3 𝑥𝑥(𝑥 = 𝐴𝜑)
3 ceqex 2879 . . 3 (𝑥 = 𝐴 → (𝜑 ↔ ∃𝑥(𝑥 = 𝐴𝜑)))
42, 3ceqsalg 2780 . 2 (𝐴𝑉 → (∀𝑥(𝑥 = 𝐴𝜑) ↔ ∃𝑥(𝑥 = 𝐴𝜑)))
51, 4bitr4id 199 1 (𝐴𝑉 → ([𝐴 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝐴𝜑)))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 104  wb 105  wal 1362   = wceq 1364  wex 1503  wcel 2160  [wsbc 2977
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2171
This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1472  df-sb 1774  df-clab 2176  df-cleq 2182  df-clel 2185  df-nfc 2321  df-v 2754  df-sbc 2978
This theorem is referenced by:  sbc6  3003  sbciegft  3008  ralsnsg  3644  ralsns  3645  fz1sbc  10125
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