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Theorem equs45f 1680
 Description: Two ways of expressing substitution when y is not free in φ. (Contributed by NM, 25-Apr-2008.)
Hypothesis
Ref Expression
equs45f.1 (φyφ)
Assertion
Ref Expression
equs45f (x(x = y φ) ↔ x(x = yφ))

Proof of Theorem equs45f
StepHypRef Expression
1 equs45f.1 . . . . 5 (φyφ)
21anim2i 324 . . . 4 ((x = y φ) → (x = y yφ))
32eximi 1488 . . 3 (x(x = y φ) → x(x = y yφ))
4 equs5a 1672 . . 3 (x(x = y yφ) → x(x = yφ))
53, 4syl 14 . 2 (x(x = y φ) → x(x = yφ))
6 equs4 1610 . 2 (x(x = yφ) → x(x = y φ))
75, 6impbii 117 1 (x(x = y φ) ↔ x(x = yφ))
 Colors of variables: wff set class Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98  ∀wal 1240  ∃wex 1378 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1333  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-11 1394  ax-4 1397  ax-i9 1420  ax-ial 1424 This theorem depends on definitions:  df-bi 110 This theorem is referenced by:  sb5f  1682
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