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Theorem dfsb1 2499
 Description: Alternate definition of substitution. Remark 9.1 in [Megill] p. 447 (p. 15 of the preprint). This was the original definition before df-sb 2070. Note that it does not require dummy variables in its definiens; this is done by having 𝑥 free in the first conjunct and bound in the second. Usage of this theorem is discouraged because it depends on ax-13 2379. (Contributed by BJ, 9-Jul-2023.) Revise df-sb 2070. (Revised by Wolf Lammen, 29-Jul-2023.) (New usage is discouraged.)
Assertion
Ref Expression
dfsb1 ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)))

Proof of Theorem dfsb1
StepHypRef Expression
1 sbequ2 2247 . . . 4 (𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑𝜑))
21com12 32 . . 3 ([𝑦 / 𝑥]𝜑 → (𝑥 = 𝑦𝜑))
3 sb1 2492 . . 3 ([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦𝜑))
42, 3jca 515 . 2 ([𝑦 / 𝑥]𝜑 → ((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)))
5 id 22 . . . . . 6 (𝑥 = 𝑦𝑥 = 𝑦)
6 sbequ1 2246 . . . . . 6 (𝑥 = 𝑦 → (𝜑 → [𝑦 / 𝑥]𝜑))
75, 6embantd 59 . . . . 5 (𝑥 = 𝑦 → ((𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑))
87sps 2182 . . . 4 (∀𝑥 𝑥 = 𝑦 → ((𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑))
98adantrd 495 . . 3 (∀𝑥 𝑥 = 𝑦 → (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) → [𝑦 / 𝑥]𝜑))
10 sb3 2491 . . . 4 (¬ ∀𝑥 𝑥 = 𝑦 → (∃𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑))
1110adantld 494 . . 3 (¬ ∀𝑥 𝑥 = 𝑦 → (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) → [𝑦 / 𝑥]𝜑))
129, 11pm2.61i 185 . 2 (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) → [𝑦 / 𝑥]𝜑)
134, 12impbii 212 1 ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 399  ∀wal 1536  ∃wex 1781  [wsb 2069 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-10 2142  ax-12 2175  ax-13 2379 This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-ex 1782  df-nf 1786  df-sb 2070 This theorem is referenced by:  drsb1  2513  subsym1  33885  bj-dfsb2  34273  frege55b  40593
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