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Theorem sbelx 1991
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 1520 . 2 (𝜑 → ∀𝑥𝜑)
21sb5rf 1846 1 (𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ [𝑥 / 𝑦]𝜑))
Colors of variables: wff set class
Syntax hints:  wa 103  wb 104  wex 1486  [wsb 1756
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1441  ax-gen 1443  ax-ie1 1487  ax-ie2 1488  ax-8 1498  ax-11 1500  ax-4 1504  ax-17 1520  ax-i9 1524  ax-ial 1528
This theorem depends on definitions:  df-bi 116  df-sb 1757
This theorem is referenced by:  sbel2x  1992
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