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Theorem sb5f 2040
 Description: Equivalence for substitution when y is not free in φ. (Contributed by NM, 5-Aug-1993.) (Revised by Mario Carneiro, 4-Oct-2016.)
Hypothesis
Ref Expression
sb6f.1 yφ
Assertion
Ref Expression
sb5f ([y / x]φx(x = y φ))

Proof of Theorem sb5f
StepHypRef Expression
1 sb6f.1 . . 3 yφ
21sb6f 2039 . 2 ([y / x]φx(x = yφ))
31equs45f 1989 . 2 (x(x = y φ) ↔ x(x = yφ))
42, 3bitr4i 243 1 ([y / x]φx(x = y φ))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 176   ∧ wa 358  ∀wal 1540  ∃wex 1541  Ⅎwnf 1544  [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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