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Theorem nfbid 1820
 Description: If in a context 𝑥 is not free in 𝜓 and 𝜒, it is not free in (𝜓 ↔ 𝜒). (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 658 . 2 ((𝜓𝜒) ↔ ((𝜓𝜒) ∧ (𝜒𝜓)))
2 nfbid.1 . . . 4 (𝜑 → Ⅎ𝑥𝜓)
3 nfbid.2 . . . 4 (𝜑 → Ⅎ𝑥𝜒)
42, 3nfimd 1812 . . 3 (𝜑 → Ⅎ𝑥(𝜓𝜒))
53, 2nfimd 1812 . . 3 (𝜑 → Ⅎ𝑥(𝜒𝜓))
64, 5nfand 1814 . 2 (𝜑 → Ⅎ𝑥((𝜓𝜒) ∧ (𝜒𝜓)))
71, 6nfxfrd 1772 1 (𝜑 → Ⅎ𝑥(𝜓𝜒))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 195   ∧ wa 383  Ⅎwnf 1699 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728 This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-ex 1696  df-nf 1701 This theorem is referenced by:  nfbi  1821  nfeud2  2470  nfeqd  2758  nfiotad  5771  iota2df  5792  axextnd  9292  axrepndlem1  9293  axrepndlem2  9294  axacndlem4  9311  axacndlem5  9312  axacnd  9313  axextdist  30949  wl-eudf  32533  wl-sb8eut  32538
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