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Theorem sbelx 1889
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 1435 . 2 (𝜑 → ∀𝑥𝜑)
21sb5rf 1748 1 (𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ [𝑥 / 𝑦]𝜑))
Colors of variables: wff set class
Syntax hints:  wa 101  wb 102  wex 1397  [wsb 1661
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-5 1352  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-8 1411  ax-11 1413  ax-4 1416  ax-17 1435  ax-i9 1439  ax-ial 1443
This theorem depends on definitions:  df-bi 114  df-sb 1662
This theorem is referenced by:  sbel2x  1890
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