Mathbox for BJ < Previous   Next > Nearby theorems Mirrors  >  Home  >  MPE Home  >  Th. List  >   Mathboxes  >  bj-sb6 Structured version   Visualization version   GIF version

Theorem bj-sb6 31955
 Description: Remove dependency on ax-13 2234 from sb6 2417. (Contributed by BJ, 11-Sep-2019.) (Proof modification is discouraged.)
Assertion
Ref Expression
bj-sb6 ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦𝜑))
Distinct variable group:   𝑥,𝑦
Allowed substitution hints:   𝜑(𝑥,𝑦)

Proof of Theorem bj-sb6
StepHypRef Expression
1 sb1 1870 . . 3 ([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦𝜑))
2 sb56 2136 . . 3 (∃𝑥(𝑥 = 𝑦𝜑) ↔ ∀𝑥(𝑥 = 𝑦𝜑))
31, 2sylib 207 . 2 ([𝑦 / 𝑥]𝜑 → ∀𝑥(𝑥 = 𝑦𝜑))
4 bj-sb2v 31941 . 2 (∀𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑)
53, 4impbii 198 1 ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦𝜑))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 195   ∧ wa 383  ∀wal 1473  ∃wex 1695  [wsb 1867 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-12 2034 This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-ex 1696  df-nf 1701  df-sb 1868 This theorem is referenced by:  bj-sb5  31956
 Copyright terms: Public domain W3C validator