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Theorem sbelx 2008
Description: Elimination of substitution. (Contributed by NM, 5-Aug-1993.)
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 ax-17 1536 . 2 (𝜑 → ∀𝑥𝜑)
21sb5rf 1862 1 (𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ [𝑥 / 𝑦]𝜑))
Colors of variables: wff set class
Syntax hints:  wa 104  wb 105  wex 1502  [wsb 1772
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1457  ax-gen 1459  ax-ie1 1503  ax-ie2 1504  ax-8 1514  ax-11 1516  ax-4 1520  ax-17 1536  ax-i9 1540  ax-ial 1544
This theorem depends on definitions:  df-bi 117  df-sb 1773
This theorem is referenced by:  sbel2x  2009
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