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Theorem nfd 1413
Description: Deduce that x is not free in ψ in a context. (Contributed by Mario Carneiro, 24-Sep-2016.)
Hypotheses
Ref Expression
nfd.1 xφ
nfd.2 (φ → (ψxψ))
Assertion
Ref Expression
nfd (φ → Ⅎxψ)

Proof of Theorem nfd
StepHypRef Expression
1 nfd.1 . . . 4 xφ
21nfri 1409 . . 3 (φxφ)
3 nfd.2 . . 3 (φ → (ψxψ))
42, 3alrimih 1355 . 2 (φx(ψxψ))
5 df-nf 1347 . 2 (Ⅎxψx(ψxψ))
64, 5sylibr 137 1 (φ → Ⅎxψ)
Colors of variables: wff set class
Syntax hints:  wi 4  wal 1240  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-5 1333  ax-gen 1335  ax-4 1397
This theorem depends on definitions:  df-bi 110  df-nf 1347
This theorem is referenced by:  nfdh  1414  nfrimi  1415  nfnt  1543  cbv1h  1630  nfald  1640  a16nf  1743  dvelimALT  1883  dvelimfv  1884  nfsb4t  1887  hbeud  1919
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