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Theorem nfand 1866
Description: If in a context 𝑥 is not free in 𝜓 and 𝜒, it is not free in (𝜓𝜒). (Contributed by Mario Carneiro, 7-Oct-2016.)
Hypotheses
Ref Expression
nfand.1 (𝜑 → Ⅎ𝑥𝜓)
nfand.2 (𝜑 → Ⅎ𝑥𝜒)
Assertion
Ref Expression
nfand (𝜑 → Ⅎ𝑥(𝜓𝜒))

Proof of Theorem nfand
StepHypRef Expression
1 df-an 385 . 2 ((𝜓𝜒) ↔ ¬ (𝜓 → ¬ 𝜒))
2 nfand.1 . . . 4 (𝜑 → Ⅎ𝑥𝜓)
3 nfand.2 . . . . 5 (𝜑 → Ⅎ𝑥𝜒)
43nfnd 1825 . . . 4 (𝜑 → Ⅎ𝑥 ¬ 𝜒)
52, 4nfimd 1863 . . 3 (𝜑 → Ⅎ𝑥(𝜓 → ¬ 𝜒))
65nfnd 1825 . 2 (𝜑 → Ⅎ𝑥 ¬ (𝜓 → ¬ 𝜒))
71, 6nfxfrd 1820 1 (𝜑 → Ⅎ𝑥(𝜓𝜒))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 383  wnf 1748
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1762  ax-4 1777
This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-ex 1745  df-nf 1750
This theorem is referenced by:  nf3and  1867  nfan  1868  nfbid  1872  nfeld  2802  nfreud  3141  nfrmod  3142  nfrmo  3144  nfrab  3153  nfifd  4147  nfdisj  4664  dfid3  5054  nfriotad  6659  axrepndlem1  9452  axrepndlem2  9453  axunndlem1  9455  axunnd  9456  axregndlem2  9463  axinfndlem1  9465  axinfnd  9466  axacndlem4  9470  axacndlem5  9471  axacnd  9472  riotasv2d  34561
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