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Theorem nford 1546
 Description: If in a context is not free in and , it is not free in . (Contributed by Jim Kingdon, 29-Oct-2019.)
Hypotheses
Ref Expression
nford.1
nford.2
Assertion
Ref Expression
nford

Proof of Theorem nford
StepHypRef Expression
1 nford.1 . . . . 5
2 nford.2 . . . . 5
3 df-nf 1437 . . . . . . 7
4 df-nf 1437 . . . . . . 7
53, 4anbi12i 455 . . . . . 6
65biimpi 119 . . . . 5
71, 2, 6syl2anc 408 . . . 4
8 19.26 1457 . . . 4
97, 8sylibr 133 . . 3
10 orc 701 . . . . . . 7
1110alimi 1431 . . . . . 6
1211imim2i 12 . . . . 5
13 olc 700 . . . . . . 7
1413alimi 1431 . . . . . 6
1514imim2i 12 . . . . 5
1612, 15jaao 708 . . . 4
1716alimi 1431 . . 3
189, 17syl 14 . 2
19 df-nf 1437 . 2
2018, 19sylibr 133 1
 Colors of variables: wff set class Syntax hints:   wi 4   wa 103   wo 697  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-io 698  ax-5 1423  ax-gen 1425 This theorem depends on definitions:  df-bi 116  df-nf 1437 This theorem is referenced by:  nfifd  3499
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