Theorem nfand 1901

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 1862	. . . 4 ⊢ (𝜑 → Ⅎ𝑥 ¬ 𝜒)
5	2, 4	nfimd 1898	. . 3 ⊢ (𝜑 → Ⅎ𝑥(𝜓 → ¬ 𝜒))
6	5	nfnd 1862	. 2 ⊢ (𝜑 → Ⅎ𝑥 ¬ (𝜓 → ¬ 𝜒))
7	1, 6	nfxfrd 1857	1 ⊢ (𝜑 → Ⅎ𝑥(𝜓 ∧ 𝜒))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 395  Ⅎwnf 1787

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-ex 1784  df-nf 1788

This theorem is referenced by:  nf3and  1902  nfan  1903  nfbid  1906  nfeud2  2590  nfeudw  2591  nfeld  2917  nfreud  3298  nfrmod  3299  nfreuw  3300  nfrmow  3301  nfrmo  3303  nfrabw  3311  nfrab  3312  nfifd  4485  nfdisjw  5047  nfdisj  5048  nfopabd  5138  dfid3  5483  nfriotadw  7220  nfriotad  7224  axrepndlem1  10279  axrepndlem2  10280  axunndlem1  10282  axunnd  10283  axregndlem2  10290  axinfndlem1  10292  axinfnd  10293  axacndlem4  10297  axacndlem5  10298  axacnd  10299  bj-gabima  35055  cbvreud  35471  riotasv2d  36898