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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  wnf 1544  [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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