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Theorem sb3an 2070
 Description: Conjunction inside and outside of a substitution are equivalent. (Contributed by NM, 14-Dec-2006.)
Assertion
Ref Expression
sb3an ([y / x](φ ψ χ) ↔ ([y / x]φ [y / x]ψ [y / x]χ))

Proof of Theorem sb3an
StepHypRef Expression
1 df-3an 936 . . 3 ((φ ψ χ) ↔ ((φ ψ) χ))
21sbbii 1653 . 2 ([y / x](φ ψ χ) ↔ [y / x]((φ ψ) χ))
3 sban 2069 . 2 ([y / x]((φ ψ) χ) ↔ ([y / x](φ ψ) [y / x]χ))
4 sban 2069 . . . 4 ([y / x](φ ψ) ↔ ([y / x]φ [y / x]ψ))
54anbi1i 676 . . 3 (([y / x](φ ψ) [y / x]χ) ↔ (([y / x]φ [y / x]ψ) [y / x]χ))
6 df-3an 936 . . 3 (([y / x]φ [y / x]ψ [y / x]χ) ↔ (([y / x]φ [y / x]ψ) [y / x]χ))
75, 6bitr4i 243 . 2 (([y / x](φ ψ) [y / x]χ) ↔ ([y / x]φ [y / x]ψ [y / x]χ))
82, 3, 73bitri 262 1 ([y / x](φ ψ χ) ↔ ([y / x]φ [y / x]ψ [y / x]χ))
 Colors of variables: wff setvar class Syntax hints:   ↔ wb 176   ∧ wa 358   ∧ w3a 934  [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-3an 936  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649 This theorem is referenced by:  sbc3ang  3104
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