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Theorem sb6 2099
 Description: Equivalence for substitution. Compare Theorem 6.2 of [Quine] p. 40. Also proved as Lemmas 16 and 17 of [Tarski] p. 70. (Contributed by NM, 18-Aug-1993.)
Assertion
Ref Expression
sb6 ([y / x]φx(x = yφ))
Distinct variable group:   x,y
Allowed substitution hints:   φ(x,y)

Proof of Theorem sb6
StepHypRef Expression
1 sb56 2098 . . 3 (x(x = y φ) ↔ x(x = yφ))
21anbi2i 675 . 2 (((x = yφ) x(x = y φ)) ↔ ((x = yφ) x(x = yφ)))
3 df-sb 1649 . 2 ([y / x]φ ↔ ((x = yφ) x(x = y φ)))
4 sp 1747 . . 3 (x(x = yφ) → (x = yφ))
54pm4.71ri 614 . 2 (x(x = yφ) ↔ ((x = yφ) x(x = yφ)))
62, 3, 53bitr4i 268 1 ([y / x]φx(x = yφ))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 176   ∧ wa 358  ∀wal 1540  ∃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-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:  sb5  2100  2sb6  2113  sb6a  2116  exsbOLD  2131  sbal2  2134
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