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Theorem nfand 1460
 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
nfand.2
Assertion
Ref Expression
nfand

Proof of Theorem nfand
StepHypRef Expression
1 nfand.1 . . . 4
2 nfand.2 . . . 4
31, 2jca 290 . . 3
4 df-nf 1350 . . . . . 6
5 df-nf 1350 . . . . . 6
64, 5anbi12i 433 . . . . 5
7 19.26 1370 . . . . 5
86, 7bitr4i 176 . . . 4
9 prth 326 . . . . . 6
10 19.26 1370 . . . . . 6
119, 10syl6ibr 151 . . . . 5
1211alimi 1344 . . . 4
138, 12sylbi 114 . . 3
143, 13syl 14 . 2
15 df-nf 1350 . 2
1614, 15sylibr 137 1
 Colors of variables: wff set class Syntax hints:   wi 4   wa 97  wal 1241  wnf 1349 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 1336  ax-gen 1338 This theorem depends on definitions:  df-bi 110  df-nf 1350 This theorem is referenced by:  nf3and  1461  nfbid  1480  nfsbxy  1818  nfsbxyt  1819  nfeld  2193  nfrexdxy  2357  nfreudxy  2483  nfifd  3355  nfriotadxy  5476  bdsepnft  10007
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