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Theorem sbequ8 2079
 Description: Elimination of equality from antecedent after substitution. (Contributed by NM, 5-Aug-1993.)
Assertion
Ref Expression
sbequ8 ([y / x]φ ↔ [y / x](x = yφ))

Proof of Theorem sbequ8
StepHypRef Expression
1 equsb1 2034 . . 3 [y / x]x = y
21a1bi 327 . 2 ([y / x]φ ↔ ([y / x]x = y → [y / x]φ))
3 sbim 2065 . 2 ([y / x](x = yφ) ↔ ([y / x]x = y → [y / x]φ))
42, 3bitr4i 243 1 ([y / x]φ ↔ [y / x](x = yφ))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 176  [wsb 1648 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925 This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649 This theorem is referenced by: (None)
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