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Theorem sbelx 2181
Description: Elimination of substitution. Also see sbel2x 2421. (Contributed by NM, 5-Aug-1993.) Avoid ax-13 2301. (Revised by Wolf Lammen, 6-Aug-2023.) Avoid ax-10 2079. (Revised by Gino Giotto, 20-Aug-2023.)
Assertion
Ref Expression
sbelx (𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ [𝑥 / 𝑦]𝜑))
Distinct variable groups:   𝑥,𝑦   𝜑,𝑥
Allowed substitution hint:   𝜑(𝑦)

Proof of Theorem sbelx
StepHypRef Expression
1 sbequ12r 2180 . . 3 (𝑥 = 𝑦 → ([𝑥 / 𝑦]𝜑𝜑))
21equsexvw 1962 . 2 (∃𝑥(𝑥 = 𝑦 ∧ [𝑥 / 𝑦]𝜑) ↔ 𝜑)
32bicomi 216 1 (𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ [𝑥 / 𝑦]𝜑))
Colors of variables: wff setvar class
Syntax hints:  wb 198  wa 387  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-12 2106
This theorem depends on definitions:  df-bi 199  df-an 388  df-ex 1743  df-sb 2016
This theorem is referenced by:  pm13.196a  40163
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