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Theorem dfsb1 2464
 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 2043. 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. (Contributed by BJ, 9-Jul-2023.) Revise df-sb 2043. (Revised by Wolf Lammen, 29-Jul-2023.)
Assertion
Ref Expression
dfsb1 ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)))

Proof of Theorem dfsb1
StepHypRef Expression
1 sbequ2 2214 . . . 4 (𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑𝜑))
21com12 32 . . 3 ([𝑦 / 𝑥]𝜑 → (𝑥 = 𝑦𝜑))
3 sb1 2461 . . 3 ([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦𝜑))
42, 3jca 512 . 2 ([𝑦 / 𝑥]𝜑 → ((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)))
5 id 22 . . . . . 6 (𝑥 = 𝑦𝑥 = 𝑦)
6 sbequ1 2213 . . . . . 6 (𝑥 = 𝑦 → (𝜑 → [𝑦 / 𝑥]𝜑))
75, 6embantd 59 . . . . 5 (𝑥 = 𝑦 → ((𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑))
87sps 2148 . . . 4 (∀𝑥 𝑥 = 𝑦 → ((𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑))
98adantrd 492 . . 3 (∀𝑥 𝑥 = 𝑦 → (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) → [𝑦 / 𝑥]𝜑))
10 sb3 2459 . . . 4 (¬ ∀𝑥 𝑥 = 𝑦 → (∃𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑))
1110adantld 491 . . 3 (¬ ∀𝑥 𝑥 = 𝑦 → (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) → [𝑦 / 𝑥]𝜑))
129, 11pm2.61i 183 . 2 (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) → [𝑦 / 𝑥]𝜑)
134, 12impbii 210 1 ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 207   ∧ wa 396  ∀wal 1520  ∃wex 1761  [wsb 2042 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1777  ax-4 1791  ax-5 1888  ax-6 1947  ax-7 1992  ax-10 2112  ax-12 2141  ax-13 2344 This theorem depends on definitions:  df-bi 208  df-an 397  df-or 843  df-ex 1762  df-nf 1766  df-sb 2043 This theorem is referenced by:  sb2vOLDOLD  2466  sbequ2OLD  2468  sbimiOLD  2469  sbimdvOLD  2470  equsb1vOLDOLD  2471  sbimdOLD  2472  sbtvOLD  2473  sbequ1OLD  2474  drsb1  2489  sbnOLD  2495  subsym1  33384  bj-dfsb2  33731  frege55b  39728
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