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Theorem nfand 1457
Description: If in a context is not free in and , it is not free in . (Contributed by Mario Carneiro, 7-Oct-2016.)
Hypotheses
Ref Expression
nfand.1  F/
nfand.2  F/
Assertion
Ref Expression
nfand  F/

Proof of Theorem nfand
StepHypRef Expression
1 nfand.1 . . . 4  F/
2 nfand.2 . . . 4  F/
31, 2jca 290 . . 3  F/  F/
4 df-nf 1347 . . . . . 6  F/
5 df-nf 1347 . . . . . 6  F/
64, 5anbi12i 433 . . . . 5  F/  F/
7 19.26 1367 . . . . 5
86, 7bitr4i 176 . . . 4  F/  F/
9 prth 326 . . . . . 6
10 19.26 1367 . . . . . 6
119, 10syl6ibr 151 . . . . 5
1211alimi 1341 . . . 4
138, 12sylbi 114 . . 3  F/  F/
143, 13syl 14 . 2
15 df-nf 1347 . 2  F/
1614, 15sylibr 137 1  F/
Colors of variables: wff set class
Syntax hints:   wi 4   wa 97  wal 1240   F/wnf 1346
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1333  ax-gen 1335
This theorem depends on definitions:  df-bi 110  df-nf 1347
This theorem is referenced by:  nf3and  1458  nfbid  1477  nfsbxy  1815  nfsbxyt  1816  nfeld  2190  nfrexdxy  2351  nfreudxy  2477  nfifd  3349  nfriotadxy  5419  bdsepnft  9321
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