Theorem nfal 1600

Description: If 𝑥 is not free in 𝜑, it is not free in ∀𝑦𝜑. (Contributed by Mario Carneiro, 11-Aug-2016.) Remove dependency on ax-4 1534. (Revised by GG, 25-Aug-2024.)

Hypothesis

Ref	Expression
nfal.1	⊢ Ⅎ𝑥𝜑

Assertion

Ref	Expression
nfal	⊢ Ⅎ𝑥∀𝑦𝜑

Proof of Theorem nfal

Step	Hyp	Ref	Expression
1		df-nf 1485	. . . . . 6 ⊢ (Ⅎ𝑥𝜑 ↔ ∀𝑥(𝜑 → ∀𝑥𝜑))
2	1	biimpi 120	. . . . 5 ⊢ (Ⅎ𝑥𝜑 → ∀𝑥(𝜑 → ∀𝑥𝜑))
3	2	alimi 1479	. . . 4 ⊢ (∀𝑦Ⅎ𝑥𝜑 → ∀𝑦∀𝑥(𝜑 → ∀𝑥𝜑))
4		ax-7 1472	. . . 4 ⊢ (∀𝑦∀𝑥(𝜑 → ∀𝑥𝜑) → ∀𝑥∀𝑦(𝜑 → ∀𝑥𝜑))
5		ax-5 1471	. . . . . 6 ⊢ (∀𝑦(𝜑 → ∀𝑥𝜑) → (∀𝑦𝜑 → ∀𝑦∀𝑥𝜑))
6		ax-7 1472	. . . . . 6 ⊢ (∀𝑦∀𝑥𝜑 → ∀𝑥∀𝑦𝜑)
7	5, 6	syl6 33	. . . . 5 ⊢ (∀𝑦(𝜑 → ∀𝑥𝜑) → (∀𝑦𝜑 → ∀𝑥∀𝑦𝜑))
8	7	alimi 1479	. . . 4 ⊢ (∀𝑥∀𝑦(𝜑 → ∀𝑥𝜑) → ∀𝑥(∀𝑦𝜑 → ∀𝑥∀𝑦𝜑))
9	3, 4, 8	3syl 17	. . 3 ⊢ (∀𝑦Ⅎ𝑥𝜑 → ∀𝑥(∀𝑦𝜑 → ∀𝑥∀𝑦𝜑))
10		df-nf 1485	. . 3 ⊢ (Ⅎ𝑥∀𝑦𝜑 ↔ ∀𝑥(∀𝑦𝜑 → ∀𝑥∀𝑦𝜑))
11	9, 10	sylibr 134	. 2 ⊢ (∀𝑦Ⅎ𝑥𝜑 → Ⅎ𝑥∀𝑦𝜑)
12		nfal.1	. 2 ⊢ Ⅎ𝑥𝜑
13	11, 12	mpg 1475	1 ⊢ Ⅎ𝑥∀𝑦𝜑

Syntax hints:   → wi 4  ∀wal 1371  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-7 1472  ax-gen 1473

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

This theorem is referenced by:  nfnf  1601  nfa2  1603  aaan  1611  cbv3  1766  cbv2  1773  nfald  1784  cbval2  1946  nfsb4t  2043  nfeuv  2073  mo23  2096  bm1.1  2191  nfnfc1  2352  nfnfc  2356  nfeq  2357  nfabdw  2368  sbcnestgf  3149  dfnfc2  3877  nfdisjv  4042  nfdisj1  4043  nffr  4409  uchoice  6241  exmidfodomrlemr  7336  exmidfodomrlemrALT  7337  exmidunben  12882  bdsepnft  15992  bdsepnfALT  15994  setindft  16070  strcollnft  16089  pw1nct  16112