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Theorem nfnf 1565
Description: If 𝑥 is not free in 𝜑, it is not free in 𝑦𝜑. (Contributed by Mario Carneiro, 11-Aug-2016.) (Proof shortened by Wolf Lammen, 30-Dec-2017.)
Hypothesis
Ref Expression
nfal.1 𝑥𝜑
Assertion
Ref Expression
nfnf 𝑥𝑦𝜑

Proof of Theorem nfnf
StepHypRef Expression
1 df-nf 1449 . 2 (Ⅎ𝑦𝜑 ↔ ∀𝑦(𝜑 → ∀𝑦𝜑))
2 nfal.1 . . . 4 𝑥𝜑
32nfal 1564 . . . 4 𝑥𝑦𝜑
42, 3nfim 1560 . . 3 𝑥(𝜑 → ∀𝑦𝜑)
54nfal 1564 . 2 𝑥𝑦(𝜑 → ∀𝑦𝜑)
61, 5nfxfr 1462 1 𝑥𝑦𝜑
Colors of variables: wff set class
Syntax hints:  wi 4  wal 1341  wnf 1448
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1435  ax-7 1436  ax-gen 1437  ax-4 1498  ax-ial 1522  ax-i5r 1523
This theorem depends on definitions:  df-bi 116  df-nf 1449
This theorem is referenced by:  nfnfc  2315
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