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Theorem sb1 2496
Description: One direction of a simplified definition of substitution. The converse requires either a disjoint variable condition (sb5 2267) or a non-freeness hypothesis (sb5f 2531). See also sb1v 2086. (Contributed by NM, 13-May-1993.) Revise df-sb 2061. (Revised by Wolf Lammen, 21-Feb-2024.)
Assertion
Ref Expression
sb1 ([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦𝜑))

Proof of Theorem sb1
StepHypRef Expression
1 spsbe 2079 . . 3 ([𝑦 / 𝑥]𝜑 → ∃𝑥𝜑)
2 pm3.2 470 . . . 4 (𝑥 = 𝑦 → (𝜑 → (𝑥 = 𝑦𝜑)))
32aleximi 1823 . . 3 (∀𝑥 𝑥 = 𝑦 → (∃𝑥𝜑 → ∃𝑥(𝑥 = 𝑦𝜑)))
41, 3syl5 34 . 2 (∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦𝜑)))
5 sb3b 2494 . . 3 (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦𝜑)))
65biimpd 230 . 2 (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦𝜑)))
74, 6pm2.61i 183 1 ([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦𝜑))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 396  wal 1526  wex 1771  [wsb 2060
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1787  ax-4 1801  ax-5 1902  ax-6 1961  ax-7 2006  ax-10 2136  ax-12 2167  ax-13 2381
This theorem depends on definitions:  df-bi 208  df-an 397  df-or 842  df-ex 1772  df-nf 1776  df-sb 2061
This theorem is referenced by:  sb3bOLD  2501  dfsb1  2503  spsbeOLDOLD  2504  sb4vOLDOLD  2506  sb4e  2517
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