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Theorem nfnf 1485
Description: If  x is not free in  ph, it is not free in  F/ y ph. (Contributed by Mario Carneiro, 11-Aug-2016.) (Proof shortened by Wolf Lammen, 30-Dec-2017.)
Hypothesis
Ref Expression
nfal.1  |-  F/ x ph
Assertion
Ref Expression
nfnf  |-  F/ x F/ y ph

Proof of Theorem nfnf
StepHypRef Expression
1 df-nf 1366 . 2  |-  ( F/ y ph  <->  A. y
( ph  ->  A. y ph ) )
2 nfal.1 . . . 4  |-  F/ x ph
32nfal 1484 . . . 4  |-  F/ x A. y ph
42, 3nfim 1480 . . 3  |-  F/ x
( ph  ->  A. y ph )
54nfal 1484 . 2  |-  F/ x A. y ( ph  ->  A. y ph )
61, 5nfxfr 1379 1  |-  F/ x F/ y ph
Colors of variables: wff set class
Syntax hints:    -> wi 4   A.wal 1257   F/wnf 1365
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-5 1352  ax-7 1353  ax-gen 1354  ax-4 1416  ax-ial 1443  ax-i5r 1444
This theorem depends on definitions:  df-bi 114  df-nf 1366
This theorem is referenced by:  nfnfc  2200
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