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Theorem nfal 1629
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 1563. (Revised by GG, 25-Aug-2024.)
Hypothesis
Ref Expression
nfal.1 𝑥𝜑
Assertion
Ref Expression
nfal 𝑥𝑦𝜑

Proof of Theorem nfal
StepHypRef Expression
1 df-nf 1514 . . . . . 6 (Ⅎ𝑥𝜑 ↔ ∀𝑥(𝜑 → ∀𝑥𝜑))
21biimpi 120 . . . . 5 (Ⅎ𝑥𝜑 → ∀𝑥(𝜑 → ∀𝑥𝜑))
32alimi 1508 . . . 4 (∀𝑦𝑥𝜑 → ∀𝑦𝑥(𝜑 → ∀𝑥𝜑))
4 ax-7 1501 . . . 4 (∀𝑦𝑥(𝜑 → ∀𝑥𝜑) → ∀𝑥𝑦(𝜑 → ∀𝑥𝜑))
5 ax-5 1500 . . . . . 6 (∀𝑦(𝜑 → ∀𝑥𝜑) → (∀𝑦𝜑 → ∀𝑦𝑥𝜑))
6 ax-7 1501 . . . . . 6 (∀𝑦𝑥𝜑 → ∀𝑥𝑦𝜑)
75, 6syl6 33 . . . . 5 (∀𝑦(𝜑 → ∀𝑥𝜑) → (∀𝑦𝜑 → ∀𝑥𝑦𝜑))
87alimi 1508 . . . 4 (∀𝑥𝑦(𝜑 → ∀𝑥𝜑) → ∀𝑥(∀𝑦𝜑 → ∀𝑥𝑦𝜑))
93, 4, 83syl 17 . . 3 (∀𝑦𝑥𝜑 → ∀𝑥(∀𝑦𝜑 → ∀𝑥𝑦𝜑))
10 df-nf 1514 . . 3 (Ⅎ𝑥𝑦𝜑 ↔ ∀𝑥(∀𝑦𝜑 → ∀𝑥𝑦𝜑))
119, 10sylibr 134 . 2 (∀𝑦𝑥𝜑 → Ⅎ𝑥𝑦𝜑)
12 nfal.1 . 2 𝑥𝜑
1311, 12mpg 1504 1 𝑥𝑦𝜑
Colors of variables:    wff set class
This proof depends on syntax axioms:  wi 4  wal 1400  wnf 1513
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1500  ax-7 1501  ax-gen 1502
This proof depends on definitions:  df-bi 117  df-nf 1514
This theorem is used by:  nfnf  1630  nfa2  1632  aaan  1640  cbv3  1795  cbv2  1802  nfald  1813  cbval2  1977  nfsb4t  2074  nfeuv  2104  mo23  2128  bm1.1  2223  nfnfc1  2395  nfnfc  2399  nfeq  2400  nfabdw  2411  sbcnestgf  3199  dfnfc2  3953  nfdisjv  4118  nfdisj1  4119  nffr  4494  uchoice  6371  modom  7108  exmidfodomrlemr  7554  exmidfodomrlemrALT  7555  exmidunben  13317  bdsepnft  16913  bdsepnfALT  16915  setindft  16991  strcollnft  17010  pw1nct  17033  nfals  17144  nfalseu  17175
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