Theorem nf3an 1498

Description: If 𝑥 is not free in 𝜑, 𝜓, and 𝜒, it is not free in (𝜑 ∧ 𝜓 ∧ 𝜒). (Contributed by Mario Carneiro, 11-Aug-2016.)

Hypotheses

Ref	Expression
nfan.1	⊢ Ⅎ𝑥𝜑
nfan.2	⊢ Ⅎ𝑥𝜓
nfan.3	⊢ Ⅎ𝑥𝜒

Assertion

Ref	Expression
nf3an	⊢ Ⅎ𝑥(𝜑 ∧ 𝜓 ∧ 𝜒)

Proof of Theorem nf3an

Step	Hyp	Ref	Expression
1		df-3an 921	. 2 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ((𝜑 ∧ 𝜓) ∧ 𝜒))
2		nfan.1	. . . 4 ⊢ Ⅎ𝑥𝜑
3		nfan.2	. . . 4 ⊢ Ⅎ𝑥𝜓
4	2, 3	nfan 1497	. . 3 ⊢ Ⅎ𝑥(𝜑 ∧ 𝜓)
5		nfan.3	. . 3 ⊢ Ⅎ𝑥𝜒
6	4, 5	nfan 1497	. 2 ⊢ Ⅎ𝑥((𝜑 ∧ 𝜓) ∧ 𝜒)
7	1, 6	nfxfr 1403	1 ⊢ Ⅎ𝑥(𝜑 ∧ 𝜓 ∧ 𝜒)

Syntax hints:   ∧ wa 102   ∧ w3a 919  Ⅎwnf 1389

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1376  ax-gen 1378  ax-4 1440

This theorem depends on definitions:  df-bi 115  df-3an 921  df-nf 1390

This theorem is referenced by:  vtocl3gaf  2667  mob  2774  nfop  3586