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Theorem nfbid 2005
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 468 . 2 ((𝜓𝜒) ↔ ((𝜓𝜒) ∧ (𝜒𝜓)))
2 nfbid.1 . . . 4 (𝜑 → Ⅎ𝑥𝜓)
3 nfbid.2 . . . 4 (𝜑 → Ⅎ𝑥𝜒)
42, 3nfimd 1996 . . 3 (𝜑 → Ⅎ𝑥(𝜓𝜒))
53, 2nfimd 1996 . . 3 (𝜑 → Ⅎ𝑥(𝜒𝜓))
64, 5nfand 2000 . 2 (𝜑 → Ⅎ𝑥((𝜓𝜒) ∧ (𝜒𝜓)))
71, 6nfxfrd 1953 1 (𝜑 → Ⅎ𝑥(𝜓𝜒))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 198  wa 386  wnf 1882
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1894  ax-4 1908
This theorem depends on definitions:  df-bi 199  df-an 387  df-or 879  df-ex 1879  df-nf 1883
This theorem is referenced by:  nfbi  2006  nfeud2OLD  2676  nfeqd  2977  nfiotad  6093  iota2df  6114  axextnd  9735  axrepndlem1  9736  axrepndlem2  9737  axacndlem4  9754  axacndlem5  9755  axacnd  9756  axextdist  32238  wl-eudf  33893  wl-sb8eut  33898
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