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Theorem bj-dfsb2 33649
 Description: Alternate (dual) definition of substitution df-sb 2016 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 2430 . 2 ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)))
2 bj-sbsb 33648 . 2 (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) ↔ (∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)))
31, 2bitri 267 1 ([𝑦 / 𝑥]𝜑 ↔ (∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 198   ∧ wa 387   ∨ wo 833  ∀wal 1505  ∃wex 1742  [wsb 2015 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1758  ax-4 1772  ax-5 1869  ax-6 1928  ax-7 1965  ax-10 2079  ax-12 2106  ax-13 2301 This theorem depends on definitions:  df-bi 199  df-an 388  df-or 834  df-ex 1743  df-nf 1747  df-sb 2016 This theorem is referenced by: (None)
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