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Theorem nfnan 2000
Description: If 𝑥 is not free in 𝜑 and 𝜓, then it is not free in (𝜑𝜓). (Contributed by Scott Fenton, 2-Jan-2018.)
Hypotheses
Ref Expression
nfan.1 𝑥𝜑
nfan.2 𝑥𝜓
Assertion
Ref Expression
nfnan 𝑥(𝜑𝜓)

Proof of Theorem nfnan
StepHypRef Expression
1 df-nan 1610 . 2 ((𝜑𝜓) ↔ ¬ (𝜑𝜓))
2 nfan.1 . . . 4 𝑥𝜑
3 nfan.2 . . . 4 𝑥𝜓
42, 3nfan 1999 . . 3 𝑥(𝜑𝜓)
54nfn 1954 . 2 𝑥 ¬ (𝜑𝜓)
61, 5nfxfr 1949 1 𝑥(𝜑𝜓)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wa 385  wnan 1609  wnf 1879
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1891  ax-4 1905
This theorem depends on definitions:  df-bi 199  df-an 386  df-or 875  df-nan 1610  df-tru 1657  df-ex 1876  df-nf 1880
This theorem is referenced by: (None)
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