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Theorem sbequ1 1918
 Description: An equality theorem for substitution. (Contributed by NM, 5-Aug-1993.)
Assertion
Ref Expression
sbequ1 (x = y → (φ → [y / x]φ))

Proof of Theorem sbequ1
StepHypRef Expression
1 pm3.4 544 . . 3 ((x = y φ) → (x = yφ))
2 19.8a 1756 . . 3 ((x = y φ) → x(x = y φ))
3 df-sb 1649 . . 3 ([y / x]φ ↔ ((x = yφ) x(x = y φ)))
41, 2, 3sylanbrc 645 . 2 ((x = y φ) → [y / x]φ)
54ex 423 1 (x = y → (φ → [y / x]φ))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ∧ wa 358  ∃wex 1541  [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-11 1746 This theorem depends on definitions:  df-bi 177  df-an 360  df-ex 1542  df-sb 1649 This theorem is referenced by:  sbequ12  1919  dfsb2  2055  sbequi  2059  sbn  2062  sbi1  2063  sb6rf  2091  mo  2226
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