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Theorem bj-dfsb2 32013
 Description: Alternate (dual) definition of substitution df-sb 1868 not using dummy variables. (Contributed by BJ, 19-Mar-2021.)
Assertion
Ref Expression
bj-dfsb2 ([𝑦 / 𝑥]𝜑 ↔ (∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)))

Proof of Theorem bj-dfsb2
StepHypRef Expression
1 df-sb 1868 . 2 ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)))
2 bj-sbsb 32012 . 2 (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) ↔ (∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)))
31, 2bitri 263 1 ([𝑦 / 𝑥]𝜑 ↔ (∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 195   ∨ wo 382   ∧ 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  ax-13 2234 This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868 This theorem is referenced by: (None)
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