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Theorem nfor 1466
 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(𝜑𝜓)

Proof of Theorem nfor
StepHypRef Expression
1 nfor.1 . . . 4 𝑥𝜑
21nfri 1412 . . 3 (𝜑 → ∀𝑥𝜑)
3 nfor.2 . . . 4 𝑥𝜓
43nfri 1412 . . 3 (𝜓 → ∀𝑥𝜓)
52, 4hbor 1438 . 2 ((𝜑𝜓) → ∀𝑥(𝜑𝜓))
65nfi 1351 1 𝑥(𝜑𝜓)
 Colors of variables: wff set class Syntax hints:   ∨ wo 629  Ⅎwnf 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-io 630  ax-5 1336  ax-gen 1338  ax-4 1400 This theorem depends on definitions:  df-bi 110  df-nf 1350 This theorem is referenced by:  nfdc  1549  nfun  3099  nfpr  3420  nfso  4039  nffrec  5982  indpi  6440  nfsum1  9875  nfsum  9876  bj-findis  10104
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