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Theorem nfbid 1829
Description: If in a context 𝑥 is not free in 𝜓 and 𝜒, 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 659 . 2 ((𝜓𝜒) ↔ ((𝜓𝜒) ∧ (𝜒𝜓)))
2 nfbid.1 . . . 4 (𝜑 → Ⅎ𝑥𝜓)
3 nfbid.2 . . . 4 (𝜑 → Ⅎ𝑥𝜒)
42, 3nfimd 1820 . . 3 (𝜑 → Ⅎ𝑥(𝜓𝜒))
53, 2nfimd 1820 . . 3 (𝜑 → Ⅎ𝑥(𝜒𝜓))
64, 5nfand 1823 . 2 (𝜑 → Ⅎ𝑥((𝜓𝜒) ∧ (𝜒𝜓)))
71, 6nfxfrd 1777 1 (𝜑 → Ⅎ𝑥(𝜓𝜒))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 196  wa 384  wnf 1705
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-ex 1702  df-nf 1707
This theorem is referenced by:  nfbi  1830  nfeud2  2481  nfeqd  2768  nfiotad  5818  iota2df  5839  axextnd  9365  axrepndlem1  9366  axrepndlem2  9367  axacndlem4  9384  axacndlem5  9385  axacnd  9386  axextdist  31441  wl-eudf  33021  wl-sb8eut  33026
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