Theorem nfand 1894

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 399	. 2 ⊢ ((𝜓 ∧ 𝜒) ↔ ¬ (𝜓 → ¬ 𝜒))
2		nfand.1	. . . 4 ⊢ (𝜑 → Ⅎ𝑥𝜓)
3		nfand.2	. . . . 5 ⊢ (𝜑 → Ⅎ𝑥𝜒)
4	3	nfnd 1854	. . . 4 ⊢ (𝜑 → Ⅎ𝑥 ¬ 𝜒)
5	2, 4	nfimd 1891	. . 3 ⊢ (𝜑 → Ⅎ𝑥(𝜓 → ¬ 𝜒))
6	5	nfnd 1854	. 2 ⊢ (𝜑 → Ⅎ𝑥 ¬ (𝜓 → ¬ 𝜒))
7	1, 6	nfxfrd 1850	1 ⊢ (𝜑 → Ⅎ𝑥(𝜓 ∧ 𝜒))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 398  Ⅎwnf 1780

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-ex 1777  df-nf 1781

This theorem is referenced by:  nf3and  1895  nfan  1896  nfbid  1899  nfeud2  2672  nfeudw  2673  nfeld  2989  nfreud  3372  nfrmod  3373  nfreuw  3374  nfrmow  3375  nfrmo  3377  nfrabw  3385  nfrab  3386  nfifd  4494  nfdisjw  5035  nfdisj  5036  dfid3  5456  nfriotadw  7116  nfriotad  7119  axrepndlem1  10008  axrepndlem2  10009  axunndlem1  10011  axunnd  10012  axregndlem2  10019  axinfndlem1  10021  axinfnd  10022  axacndlem4  10026  axacndlem5  10027  axacnd  10028  cbvreud  34648  riotasv2d  36087