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Theorem nf3an 1455
 Description: If x is not free in φ, ψ, and χ, it is not free in (φ ∧ ψ ∧ χ). (Contributed by Mario Carneiro, 11-Aug-2016.)
Hypotheses
Ref Expression
nfan.1 xφ
nfan.2 xψ
nfan.3 xχ
Assertion
Ref Expression
nf3an x(φ ψ χ)

Proof of Theorem nf3an
StepHypRef Expression
1 df-3an 886 . 2 ((φ ψ χ) ↔ ((φ ψ) χ))
2 nfan.1 . . . 4 xφ
3 nfan.2 . . . 4 xψ
42, 3nfan 1454 . . 3 x(φ ψ)
5 nfan.3 . . 3 xχ
64, 5nfan 1454 . 2 x((φ ψ) χ)
71, 6nfxfr 1360 1 x(φ ψ χ)
 Colors of variables: wff set class Syntax hints:   ∧ wa 97   ∧ w3a 884  Ⅎwnf 1346 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-4 1397 This theorem depends on definitions:  df-bi 110  df-3an 886  df-nf 1347 This theorem is referenced by:  vtocl3gaf  2616  mob  2717  nfop  3555
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