Theorem nfand 1898

Description: If in a context 𝑥 is not free in 𝜓 and 𝜒, then it is not free in (𝜓 ∧ 𝜒). (Contributed by Mario Carneiro, 7-Oct-2016.)

Hypotheses

Ref	Expression
nfand.1	⊢ (𝜑 → Ⅎ𝑥𝜓)
nfand.2	⊢ (𝜑 → Ⅎ𝑥𝜒)

Assertion

Ref	Expression
nfand	⊢ (𝜑 → Ⅎ𝑥(𝜓 ∧ 𝜒))

Proof of Theorem nfand

Step	Hyp	Ref	Expression
1		df-an 400	. 2 ⊢ ((𝜓 ∧ 𝜒) ↔ ¬ (𝜓 → ¬ 𝜒))
2		nfand.1	. . . 4 ⊢ (𝜑 → Ⅎ𝑥𝜓)
3		nfand.2	. . . . 5 ⊢ (𝜑 → Ⅎ𝑥𝜒)
4	3	nfnd 1859	. . . 4 ⊢ (𝜑 → Ⅎ𝑥 ¬ 𝜒)
5	2, 4	nfimd 1895	. . 3 ⊢ (𝜑 → Ⅎ𝑥(𝜓 → ¬ 𝜒))
6	5	nfnd 1859	. 2 ⊢ (𝜑 → Ⅎ𝑥 ¬ (𝜓 → ¬ 𝜒))
7	1, 6	nfxfrd 1855	1 ⊢ (𝜑 → Ⅎ𝑥(𝜓 ∧ 𝜒))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 399  Ⅎwnf 1785

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-ex 1782  df-nf 1786

This theorem is referenced by:  nf3and  1899  nfan  1900  nfbid  1903  nfeud2  2651  nfeudw  2652  nfeld  2966  nfreud  3325  nfrmod  3326  nfreuw  3327  nfrmow  3328  nfrmo  3330  nfrabw  3338  nfrab  3339  nfifd  4453  nfdisjw  5007  nfdisj  5008  dfid3  5427  nfriotadw  7101  nfriotad  7104  axrepndlem1  10003  axrepndlem2  10004  axunndlem1  10006  axunnd  10007  axregndlem2  10014  axinfndlem1  10016  axinfnd  10017  axacndlem4  10021  axacndlem5  10022  axacnd  10023  cbvreud  34790  riotasv2d  36253