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Theorem nfbid 1896
Description: If in a context 𝑥 is not free in 𝜓 and 𝜒, then it is not free in (𝜓𝜒). (Contributed by Mario Carneiro, 24-Sep-2016.) (Proof shortened by Wolf Lammen, 29-Dec-2017.)
Hypotheses
Ref Expression
nfbid.1 (𝜑 → Ⅎ𝑥𝜓)
nfbid.2 (𝜑 → Ⅎ𝑥𝜒)
Assertion
Ref Expression
nfbid (𝜑 → Ⅎ𝑥(𝜓𝜒))

Proof of Theorem nfbid
StepHypRef Expression
1 dfbi2 477 . 2 ((𝜓𝜒) ↔ ((𝜓𝜒) ∧ (𝜒𝜓)))
2 nfbid.1 . . . 4 (𝜑 → Ⅎ𝑥𝜓)
3 nfbid.2 . . . 4 (𝜑 → Ⅎ𝑥𝜒)
42, 3nfimd 1888 . . 3 (𝜑 → Ⅎ𝑥(𝜓𝜒))
53, 2nfimd 1888 . . 3 (𝜑 → Ⅎ𝑥(𝜒𝜓))
64, 5nfand 1891 . 2 (𝜑 → Ⅎ𝑥((𝜓𝜒) ∧ (𝜒𝜓)))
71, 6nfxfrd 1847 1 (𝜑 → Ⅎ𝑥(𝜓𝜒))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 208  wa 398  wnf 1777
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803
This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-ex 1774  df-nf 1778
This theorem is referenced by:  nfbi  1897  nfeqd  2986  nfiotadw  6310  nfiotad  6312  iota2df  6335  axextnd  10005  axrepndlem1  10006  axrepndlem2  10007  axacndlem4  10024  axacndlem5  10025  axacnd  10026  axextdist  33032  copsex2d  34417  cbveud  34635  wl-eudf  34790  wl-sb8eut  34795
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