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Theorem sbc6g 2811
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 1401 . . 3 𝑥𝑥(𝑥 = 𝐴𝜑)
2 ceqex 2694 . . 3 (𝑥 = 𝐴 → (𝜑 ↔ ∃𝑥(𝑥 = 𝐴𝜑)))
31, 2ceqsalg 2599 . 2 (𝐴𝑉 → (∀𝑥(𝑥 = 𝐴𝜑) ↔ ∃𝑥(𝑥 = 𝐴𝜑)))
4 sbc5 2810 . 2 ([𝐴 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝐴𝜑))
53, 4syl6rbbr 192 1 (𝐴𝑉 → ([𝐴 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝐴𝜑)))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 101  wb 102  wal 1257   = wceq 1259  wex 1397  wcel 1409  [wsbc 2787
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-io 640  ax-5 1352  ax-7 1353  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-8 1411  ax-10 1412  ax-11 1413  ax-i12 1414  ax-bndl 1415  ax-4 1416  ax-17 1435  ax-i9 1439  ax-ial 1443  ax-i5r 1444  ax-ext 2038
This theorem depends on definitions:  df-bi 114  df-tru 1262  df-nf 1366  df-sb 1662  df-clab 2043  df-cleq 2049  df-clel 2052  df-nfc 2183  df-v 2576  df-sbc 2788
This theorem is referenced by:  sbc6  2812  sbciegft  2816  ralsnsg  3435  ralsns  3436  fz1sbc  9060
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