Theorem nf3an 1590

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 983	. 2 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ((𝜑 ∧ 𝜓) ∧ 𝜒))
2		nfan.1	. . . 4 ⊢ Ⅎ𝑥𝜑
3		nfan.2	. . . 4 ⊢ Ⅎ𝑥𝜓
4	2, 3	nfan 1589	. . 3 ⊢ Ⅎ𝑥(𝜑 ∧ 𝜓)
5		nfan.3	. . 3 ⊢ Ⅎ𝑥𝜒
6	4, 5	nfan 1589	. 2 ⊢ Ⅎ𝑥((𝜑 ∧ 𝜓) ∧ 𝜒)
7	1, 6	nfxfr 1498	1 ⊢ Ⅎ𝑥(𝜑 ∧ 𝜓 ∧ 𝜒)

Syntax hints:   ∧ wa 104   ∧ w3a 981  Ⅎwnf 1484

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1471  ax-gen 1473  ax-4 1534

This theorem depends on definitions:  df-bi 117  df-3an 983  df-nf 1485

This theorem is referenced by:  vtocl3gaf  2844  mob  2959  nfop  3841  mkvprop  7275  seq3f1olemstep  10681  seq3f1olemp  10682  nfsum1  11742  nfsum  11743  dfgrp3mlem  13505