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Theorem sb3an 1829
 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 886 . . 3 ((φ ψ χ) ↔ ((φ ψ) χ))
21sbbii 1645 . 2 ([y / x](φ ψ χ) ↔ [y / x]((φ ψ) χ))
3 sban 1826 . 2 ([y / x]((φ ψ) χ) ↔ ([y / x](φ ψ) [y / x]χ))
4 sban 1826 . . . 4 ([y / x](φ ψ) ↔ ([y / x]φ [y / x]ψ))
54anbi1i 431 . . 3 (([y / x](φ ψ) [y / x]χ) ↔ (([y / x]φ [y / x]ψ) [y / x]χ))
6 df-3an 886 . . 3 (([y / x]φ [y / x]ψ [y / x]χ) ↔ (([y / x]φ [y / x]ψ) [y / x]χ))
75, 6bitr4i 176 . 2 (([y / x](φ ψ) [y / x]χ) ↔ ([y / x]φ [y / x]ψ [y / x]χ))
82, 3, 73bitri 195 1 ([y / x](φ ψ χ) ↔ ([y / x]φ [y / x]ψ [y / x]χ))
 Colors of variables: wff set class Syntax hints:   ∧ wa 97   ↔ wb 98   ∧ w3a 884  [wsb 1642 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-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425 This theorem depends on definitions:  df-bi 110  df-3an 886  df-nf 1347  df-sb 1643 This theorem is referenced by:  sbc3ang  2814
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