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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  3392  nfreud  3393  nfrmo  3394  nfrab  3436  nfifd  4508  nfdisjw  5074  nfdisj  5075  nfopabd  5163  dfid3  5521  nfriotadw  7318  nfriotad  7321  axrepndlem1  10505  axrepndlem2  10506  axunndlem1  10508  axunnd  10509  axregndlem2  10516  axinfndlem1  10518  axinfnd  10519  axacndlem4  10523  axacndlem5  10524  axacnd  10525  axsepg2  35051  axsepg2ALT  35052  bj-gabima  36916  cbvreud  37349  riotasv2d  38938
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