Theorem nfand 1899

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 396	. 2 ⊢ ((𝜓 ∧ 𝜒) ↔ ¬ (𝜓 → ¬ 𝜒))
2		nfand.1	. . . 4 ⊢ (𝜑 → Ⅎ𝑥𝜓)
3		nfand.2	. . . . 5 ⊢ (𝜑 → Ⅎ𝑥𝜒)
4	3	nfnd 1860	. . . 4 ⊢ (𝜑 → Ⅎ𝑥 ¬ 𝜒)
5	2, 4	nfimd 1896	. . 3 ⊢ (𝜑 → Ⅎ𝑥(𝜓 → ¬ 𝜒))
6	5	nfnd 1860	. 2 ⊢ (𝜑 → Ⅎ𝑥 ¬ (𝜓 → ¬ 𝜒))
7	1, 6	nfxfrd 1856	1 ⊢ (𝜑 → Ⅎ𝑥(𝜓 ∧ 𝜒))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 395  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 207  df-an 396  df-or 849  df-ex 1782  df-nf 1786

This theorem is referenced by:  nf3and  1900  nfan  1901  nfbid  1904  nfeud2  2591  nfeudw  2592  nfeld  2911  nfrmod  3386  nfreud  3387  nfrmo  3388  nfrab  3428  nfifd  4497  nfdisjw  5065  nfdisj  5066  nfopabd  5154  dfid3  5522  nfriotadw  7325  nfriotad  7328  axrepndlem1  10506  axrepndlem2  10507  axunndlem1  10509  axunnd  10510  axregndlem2  10517  axinfndlem1  10519  axinfnd  10520  axacndlem4  10524  axacndlem5  10525  axacnd  10526  nfchnd  18568  axsepg2  35241  axsepg2ALT  35242  axtcond  36676  bj-gabima  37263  cbvreud  37703  riotasv2d  39417