Theorem nfnf 1600

Description: If 𝑥 is not free in 𝜑, it is not free in Ⅎ𝑦𝜑. (Contributed by Mario Carneiro, 11-Aug-2016.) (Proof shortened by Wolf Lammen, 30-Dec-2017.)

Hypothesis

Ref	Expression
nfal.1	⊢ Ⅎ𝑥𝜑

Assertion

Ref	Expression
nfnf	⊢ Ⅎ𝑥Ⅎ𝑦𝜑

Proof of Theorem nfnf

Step	Hyp	Ref	Expression
1		df-nf 1484	. 2 ⊢ (Ⅎ𝑦𝜑 ↔ ∀𝑦(𝜑 → ∀𝑦𝜑))
2		nfal.1	. . . 4 ⊢ Ⅎ𝑥𝜑
3	2	nfal 1599	. . . 4 ⊢ Ⅎ𝑥∀𝑦𝜑
4	2, 3	nfim 1595	. . 3 ⊢ Ⅎ𝑥(𝜑 → ∀𝑦𝜑)
5	4	nfal 1599	. 2 ⊢ Ⅎ𝑥∀𝑦(𝜑 → ∀𝑦𝜑)
6	1, 5	nfxfr 1497	1 ⊢ Ⅎ𝑥Ⅎ𝑦𝜑

Syntax hints:   → wi 4  ∀wal 1371  Ⅎwnf 1483

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 1470  ax-7 1471  ax-gen 1472  ax-4 1533  ax-ial 1557  ax-i5r 1558

This theorem depends on definitions:  df-bi 117  df-nf 1484

This theorem is referenced by:  nfnfc  2355