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Theorem sb6a 2116
 Description: Equivalence for substitution. (Contributed by NM, 5-Aug-1993.)
Assertion
Ref Expression
sb6a ([y / x]φx(x = y → [x / y]φ))
Distinct variable group:   x,y
Allowed substitution hints:   φ(x,y)

Proof of Theorem sb6a
StepHypRef Expression
1 sb6 2099 . 2 ([y / x]φx(x = yφ))
2 sbequ12 1919 . . . . 5 (y = x → (φ ↔ [x / y]φ))
32equcoms 1681 . . . 4 (x = y → (φ ↔ [x / y]φ))
43pm5.74i 236 . . 3 ((x = yφ) ↔ (x = y → [x / y]φ))
54albii 1566 . 2 (x(x = yφ) ↔ x(x = y → [x / y]φ))
61, 5bitri 240 1 ([y / x]φx(x = y → [x / y]φ))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 176  ∀wal 1540  [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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