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Theorem nfimd 1564
 Description: If in a context is not free in and , then it is not free in . (Contributed by Mario Carneiro, 24-Sep-2016.) (Proof shortened by Wolf Lammen, 30-Dec-2017.)
Hypotheses
Ref Expression
nfimd.1
nfimd.2
Assertion
Ref Expression
nfimd

Proof of Theorem nfimd
StepHypRef Expression
1 nfimd.1 . 2
2 nfimd.2 . 2
3 nfnf1 1523 . . . . 5
43nfri 1499 . . . 4
5 nfnf1 1523 . . . . 5
65nfri 1499 . . . 4
7 nfr 1498 . . . . . 6
87imim2d 54 . . . . 5
9 19.21t 1561 . . . . . 6
109biimprd 157 . . . . 5
118, 10syl9r 73 . . . 4
124, 6, 11alrimdh 1455 . . 3
13 df-nf 1437 . . 3
1412, 13syl6ibr 161 . 2
151, 2, 14sylc 62 1
 Colors of variables: wff set class Syntax hints:   wi 4  wal 1329  wnf 1436 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 1423  ax-gen 1425  ax-4 1487  ax-ial 1514  ax-i5r 1515 This theorem depends on definitions:  df-bi 116  df-nf 1437 This theorem is referenced by:  nfbid  1567  dvelimALT  1985  dvelimfv  1986  dvelimor  1993  nfmod  2016  nfraldxy  2467  nfixpxy  6611  cbvrald  13025
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