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Theorem bj-dfsb2 36821
Description: Alternate (dual) definition of substitution df-sb 2063 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 dfsb1 2484 . 2 ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)))
2 bj-sbsb 36820 . 2 (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) ↔ (∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)))
31, 2bitri 275 1 ([𝑦 / 𝑥]𝜑 ↔ (∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206  wa 395  wo 847  wal 1535  wex 1776  [wsb 2062
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-10 2139  ax-12 2175  ax-13 2375
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-ex 1777  df-nf 1781  df-sb 2063
This theorem is referenced by: (None)
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