Theorem nfand 1822

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 360	. 2 ⊢ ((ψ ∧ χ) ↔ ¬ (ψ → ¬ χ))
2		nfand.1	. . . 4 ⊢ (φ → Ⅎxψ)
3		nfand.2	. . . . 5 ⊢ (φ → Ⅎxχ)
4	3	nfnd 1791	. . . 4 ⊢ (φ → Ⅎx ¬ χ)
5	2, 4	nfimd 1808	. . 3 ⊢ (φ → Ⅎx(ψ → ¬ χ))
6	5	nfnd 1791	. 2 ⊢ (φ → Ⅎx ¬ (ψ → ¬ χ))
7	1, 6	nfxfrd 1571	1 ⊢ (φ → Ⅎx(ψ ∧ χ))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 358  Ⅎwnf 1544

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-11 1746

This theorem depends on definitions:  df-bi 177  df-an 360  df-ex 1542  df-nf 1545

This theorem is referenced by:  nf3and  1823  nfan  1824  nfbid  1832  nfeld  2505  nfreud  2784  nfrmod  2785  nfrmo  2787  nfrab  2793  nfifd  3686  dfid3  4769