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Theorem nfand 1897
Description: If in a context 𝑥 is not free in 𝜓 and 𝜒, then 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 396 . 2 ((𝜓𝜒) ↔ ¬ (𝜓 → ¬ 𝜒))
2 nfand.1 . . . 4 (𝜑 → Ⅎ𝑥𝜓)
3 nfand.2 . . . . 5 (𝜑 → Ⅎ𝑥𝜒)
43nfnd 1858 . . . 4 (𝜑 → Ⅎ𝑥 ¬ 𝜒)
52, 4nfimd 1894 . . 3 (𝜑 → Ⅎ𝑥(𝜓 → ¬ 𝜒))
65nfnd 1858 . 2 (𝜑 → Ⅎ𝑥 ¬ (𝜓 → ¬ 𝜒))
71, 6nfxfrd 1854 1 (𝜑 → Ⅎ𝑥(𝜓𝜒))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 395  wnf 1783
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-ex 1780  df-nf 1784
This theorem is referenced by:  nf3and  1898  nfan  1899  nfbid  1902  nfeud2  2583  nfeudw  2584  nfeld  2903  nfrmod  3401  nfreud  3402  nfrmo  3403  nfrab  3445  nfifd  4518  nfdisjw  5086  nfdisj  5087  nfopabd  5175  dfid3  5536  nfriotadw  7352  nfriotad  7355  axrepndlem1  10545  axrepndlem2  10546  axunndlem1  10548  axunnd  10549  axregndlem2  10556  axinfndlem1  10558  axinfnd  10559  axacndlem4  10563  axacndlem5  10564  axacnd  10565  axsepg2  35072  axsepg2ALT  35073  bj-gabima  36928  cbvreud  37361  riotasv2d  38950
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