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Theorem nfand 1822
Description: If in a context x is not free in ψ and χ, it is not free in (ψ χ). (Contributed by Mario Carneiro, 7-Oct-2016.)
Hypotheses
Ref Expression
nfand.1 (φ → Ⅎxψ)
nfand.2 (φ → Ⅎxχ)
Assertion
Ref Expression
nfand (φ → Ⅎx(ψ χ))

Proof of Theorem nfand
StepHypRef Expression
1 df-an 360 . 2 ((ψ χ) ↔ ¬ (ψ → ¬ χ))
2 nfand.1 . . . 4 (φ → Ⅎxψ)
3 nfand.2 . . . . 5 (φ → Ⅎxχ)
43nfnd 1791 . . . 4 (φ → Ⅎx ¬ χ)
52, 4nfimd 1808 . . 3 (φ → Ⅎx(ψ → ¬ χ))
65nfnd 1791 . 2 (φ → Ⅎx ¬ (ψ → ¬ χ))
71, 6nfxfrd 1571 1 (φ → Ⅎx(ψ χ))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4   wa 358  wnf 1544
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-11 1746
This theorem depends on definitions:  df-bi 177  df-an 360  df-ex 1542  df-nf 1545
This theorem is referenced by:  nf3and  1823  nfan  1824  nfbid  1832  nfeld  2504  nfreud  2783  nfrmod  2784  nfrmo  2786  nfrab  2792  nfifd  3685  dfid3  4768
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