Theorem nfand 1866

Description: If in a context 𝑥 is not free in 𝜓 and 𝜒, 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 385	. 2 ⊢ ((𝜓 ∧ 𝜒) ↔ ¬ (𝜓 → ¬ 𝜒))
2		nfand.1	. . . 4 ⊢ (𝜑 → Ⅎ𝑥𝜓)
3		nfand.2	. . . . 5 ⊢ (𝜑 → Ⅎ𝑥𝜒)
4	3	nfnd 1825	. . . 4 ⊢ (𝜑 → Ⅎ𝑥 ¬ 𝜒)
5	2, 4	nfimd 1863	. . 3 ⊢ (𝜑 → Ⅎ𝑥(𝜓 → ¬ 𝜒))
6	5	nfnd 1825	. 2 ⊢ (𝜑 → Ⅎ𝑥 ¬ (𝜓 → ¬ 𝜒))
7	1, 6	nfxfrd 1820	1 ⊢ (𝜑 → Ⅎ𝑥(𝜓 ∧ 𝜒))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 383  Ⅎwnf 1748

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1762  ax-4 1777

This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-ex 1745  df-nf 1750

This theorem is referenced by:  nf3and  1867  nfan  1868  nfbid  1872  nfeld  2802  nfreud  3141  nfrmod  3142  nfrmo  3144  nfrab  3153  nfifd  4147  nfdisj  4664  dfid3  5054  nfriotad  6659  axrepndlem1  9452  axrepndlem2  9453  axunndlem1  9455  axunnd  9456  axregndlem2  9463  axinfndlem1  9465  axinfnd  9466  axacndlem4  9470  axacndlem5  9471  axacnd  9472  riotasv2d  34561