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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-mp 5  ax-1 6  ax-2 7  ax-3 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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