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Theorem nfor 1463
Description: If x is not free in φ and ψ, it is not free in (φ ψ). (Contributed by Jim Kingdon, 11-Mar-2018.)
Hypotheses
Ref Expression
nfor.1 xφ
nfor.2 xψ
Assertion
Ref Expression
nfor x(φ ψ)

Proof of Theorem nfor
StepHypRef Expression
1 nfor.1 . . . 4 xφ
21nfri 1409 . . 3 (φxφ)
3 nfor.2 . . . 4 xψ
43nfri 1409 . . 3 (ψxψ)
52, 4hbor 1435 . 2 ((φ ψ) → x(φ ψ))
65nfi 1348 1 x(φ ψ)
Colors of variables: wff set class
Syntax hints:   wo 628  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-io 629  ax-5 1333  ax-gen 1335  ax-4 1397
This theorem depends on definitions:  df-bi 110  df-nf 1347
This theorem is referenced by:  nfdc  1546  nfun  3093  nfpr  3411  nfso  4030  nffrec  5921  indpi  6326  bj-findis  9439
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